14.1 PERIODIC MOTION 1, What is periodic motion ? Give some of its examples. Periodic motion. Any motion that repeats itself over and over again at regular intervals of time is called periodic or harmonic motion. The smallest interval of time after which the motion is repeated is called its time period. The time period is denoted by T and its SI unit is second. Examples of periodic motion : () The motion of any planet around the sun in an elliptical orbit is periodic. The period of revolution of Mercury is 87.97 days. (ii) The motion of the moon around the earth is periodic. Its time period is 27.3 days. Gii) The motion of Halley's comet around the sun i periodic. It appears on the earth after every 76 years. (iv) The motion of the hands of a clock is periodic. (v) The heart beats of a human being are periodic. ‘The periodic time is about 08 second for a normal person. OSCILLATORY OR HARMONIC MOTION 2. What is oscillatory motion ? Give some of its examples. 14.2 OSCILLATIONS Oscillatory motion. If a body moves back and forth repeatedly about its mean position, its motion is said to be oscillatory or vibratory or harmonic motion. Such a motion repeats itself over and over again about a mean position such that it remains confined within well defined limits (known as extreme positions) on either side of the ‘mean position. Examples of oscillatory motion : (i The swinging motion of the pendulum of a wall clock. {ii) The oscillations of a mass suspended from a spring. (iii) The motion of the piston of an automobile engine. (iv) The vibrations of the string of a guitar. (v) When a freely suspended bar magnet is dis- placed from its equilibrium position along north- south line and released, it executes oscillatory motion. 14.3. PERIODIC MOTION VS. OSCILLATORY MOTION 3. Every oscillatory motion is necessarily periodic but every periodic motion need not be oscillatory. Justify. Distinction between periodic and oscillatory motions. Every oscillatory motion is necessarily periodic because it is repeated at regular intervals of 14.14.2 PHYSICS-xI time. In addition, it is bounded abdut one mean position. But every periodic motion need not be oscillatory. For example, the earth completes one revolution around the sun in 1 year but it is not a to and fro motion about any mean position. Hence its motion is periodic but not oscillatory. 14.4 © PERIODIC FUNCTIONS AND FOURIER ANALYSIS 4, With suitable examples, explain the meaning of a periodic function. Construct two infinite sets of periodic functions with period T. Hence state Fourier theorem. Periodic function. Any function that repeats itself at regular intervals of its argument is called a periodic function. Consider the function f (8) satisfying the Property, f(0+T) =F (8) This indicates that the value of the function remains same when the argument is increased or * decreased by an integral multiple of T for all values of ©. A function f satisfying this property is said to be periodic having a period T. For example, trigonometric functions like sin 6 and cos @ are periodic with a period of 27 radians, because sin (@+2n)=sin 8 cos 8 +2) = cos 0 If the independent variable 0 stands for some dimensional quantity such as time f, then we can construct periodic functions with period Tas follows: 2nt A(t)=sin 22 ana 1 ()=cos St We can check the periodicity by replacing tby t+ T. Thus 2n _((2nt T)=sin = (t+ 7) =sin( 2" 425 A(t +T)=sin = ( san + ) an 2tt asin t= (0) Similarly, g, (t+T)=8, (1) Itcan be easily seen that functions with period T /n, where n =1,2,3,....also repeat their values after a time T. Hence it is possible to construct two infinite sets of periodic functions such as nt Jf, t)=sin 2% n=1,2,3,4, 2nnt 1) = cos n=0,1,2,3,4, & (t) T In the set of cosine functions we have included the constant function ga (t)=1. The constant function 1 is periodic for any value of T and hence does not alter the periodicity of g, (#) Fourier theorem. This theorem slates that any arbitrary function F (1) with period T can be expressed as the unique combination of sine and cosine functions f(t) and g, (t) with suitable coeficients. Mathematically, it can’ be expressed as F()=by +b, cos 2! +b, cos Ey cos Rt T 2nt 4nt oat +4; sin > ay sin + ty si” ont =by + by COS wot + by COS 2eot + by cos Seat + +4, sin wf + a sin 2ot +a, sin 3at +... or F()=h+ 5b, cosnat+ La, sinnot where @=2n/T. ‘The coefficients by, By, Byy an dgy 45, are called Fourier coefficients. These coefficients can be deter- mined uniquely by a mathematical method called Fourier analysis. Suppose all the Fourier coefficients except a, and b, are zero, then F(t) ant 2nt sin TT +b cos at This equation is a special periodic motion called simple harmonic motion (S.H.M) 14.5 PERIODIC, HARMONIC AND NON-HARMONIC FUNCTIONS 5. Distinguish between periodic, harmonic and non-harmonic functions. Give examples of each. Periodic, harmonic and non-harmonic functions. Any function that repeats itself at regular intervals of its argument is called a periodic function. The following sine and cosine functions are periodic with period T : t f (t)=sin ot =sin 22! Qnt id (t)= it = == an §()= 008 of =cos Fig. 14.1. shows how these functions vary with time t. a | or & ™ Tr "ore > i (@ : t vas NE NT Ee | (be) Fig. 14.1 Periodic functions which are harmonic,OSCILLATIONS _ 14.3 Obviously, these functions vary between a maximum value + Land minimum value — 1 passing through zero in between. The periodic functions which can be represented bby a sinc or cosine curve are called harmonic functions. All harmonic functions are necessarily periodic but all periodic functions are not harmonic. The periodic functions twhiich cannot be represented by single sine or cosine function are called non-harmonic functions. Fig, 14.2 shows some periodic functions which repeat themselves in a period T but are not harmonic. 4 RO ° ( ar ar - ca rw rot 7 Fig. 14.2 Some non-harmonic periodic functions. ‘Any non-harmonic periodic function can be constructed from two or more harmonic functions. One such function is: F (#)=a, sin ot +a, sin 2at It can be easily checked that the functions tan ot and cot wt are periodic with period T = / awhile sec wt and cosec «wt are periodic with period T =2x/ Thus tan {of « + = = tan (of + )=tan ot ry wfa(t0 28) neta 2m aces But such functions take values between zero and infinity. So these functions cannot be used to represent displacement functions in periodic motions because displacement always takes a finite value in any physical situation. Exam les Based on paceman Concerts UseD 1. A function which can be represented by a single sine or cosine function is a harmonic function otherwise non-harmonic. 2. A periodic function can be expressed as the sum. of sine and cosine functions of different time periods with suitable coefficients. Examen: t. On an average a Inman heart is found to beat 75 times in a minute. Calculate its beat frequency and period. INCERTI Solution. Beat frequency of the heart, No. of beats __75 Time taken Imin 75_ 1.95 5"! =1.25 He. 60s Beat period, T=4= = 085. v 1255 amt Which of the following functions of time represent (a) periodic and (1) rion-periodic motion ? Give the period for each case of periodic motion. [ois any positive constant} INCERTI () sin ot + cos ot (ji) sin Ot + cos 2a + sin dot Win (iv) log (at). Solution. (i) Here x(t)=sin wt + cos ot = 12 [sit cos ® coe ot sin] = V2 sin (wt + 2/4) Moreover, x(1+28) a0 sin[ot+2x/0) + 0/4] ° = 2 sin( ot 6202) =P sin( ot +) x(t). Hence sin wt + cos wt is a periodic function with time period equal to2/ (ii) Here x (#)=sin ot + cos2at + sin dot sin of is a periodic function with period =2n/o=T cos 2a is a periodic function with period =2n/2o=n/o=T/2 sin 4ot is a periodic function with period =2n/4o=n/2@=T/4 Clearly, the entire function x(t) repeats after a minimum time T =21/ w. Hence the given function is periodic. (iii) The function &® decreases monotonically to zero as t—>. It is an exponential function with a negative exponent of ¢ where e=2.71828. It never repeats its value. So it is non-periodic. (iv) The function log («t) increases monotonically with time. As t > @, log (ct) > =, It never repeats its value. So it is non-periodic.14.4 PHYSICS-XI a This equation defines S.H.M. Here k is a positive ® PROBLEMS FOR PRACTICE constant called force constant or spring factor and is Which of the following fictions of time represent defined as the restoring force produced per unit displace. (a) simple harmonic motion, (b) periodic but not simple ment, The SL unit of k is Nm~!, The negative sign in the tire and (0 nor periodic notion ? Fin the period of above equation shows that the restoring force F always ech periodic motion. Here wis positive rel costent. acts in the opposite direction ofthe dleplacemment sin ef + c08 ot (Ans. Simple harmonic) Now, according to Newton’s second law of motion, 2. sin nt + 2cos 2nt + 3sin 3 ae (Ans, Periodic but not simple harmonic) ae 3. cos (20t + n/3) (Ans. Simple harmonic) k 4. sin? of, (Ans. Periodic but not simple harmonic) nm git Sy 5. cos ot + 2sin? ot Hence simple harmonic motion may also be (Ans. Periodic but not simple harmonic) defined as follows : x Hints A particle is said to possess siniple harmonic motion ifit ‘moves to and fro about a mean position under an acceleration ia) dei aoa ie which is directly proportional to its displacement from the 2. Each term Loeeys HM. mrean position and is always directed towards that position. Period of sin xt, T 7 =2s Examples of simple harmonic motion : - (i) Oscillations of a loaded spring. Period of 2cos 2nt = 2 7 tas (ii) Vibrations of a tuning fork. (iii) Vibrations of the balance wheel of a watch. (iv) Oscillations of a freely suspended magnet in a The sum is not simple harmonic but periodic with uniform magnetic field. ioe. 7. State some important features of simple harmonic 3. cos (201 + x/3) represents SH.M. with motion. T=2n/2o=n/o. 4. sin? wt = 1/2~-(1/2)cos 2ot, The function does not represent S.H.M. but is Period of 3sin 3nt Some important features of S.H.M. (i) The motion of the particle is periodic. (i It is the oscillatory motion of simplest kind in Lainie lala ty. which the particle oscillates back and forth about 5. cos at + 2sin* wt =cos wt + 1-cos 2ot its mean position with constant amplitude and = 1+ c0s wf = os 20t fixed frequency. cos wt represents SH.M. with T= 2n/a. (iii) Restoring force acting on the particle is propor- cos 2! represents S.H.M. with period tional to its displacement from the mean position, =2n/20=n/@=T/2 (iv) The force acting on the particle always opposes ‘The combined function does not represent S.H.M. the increase in its displacement. but is periodic with T= 2n/ a. (2) A simple harmonic motion can always be — ee expressed in terms of a single harmonic 14.6 SIMPLE HARMONIC MOTION function of sine or cosine. 6. Mat is meant by simple harmonic motion ? Give 447 DIFFERENTIAL EQUATION FOR S.H.M. samples. ee 8, Write down the differential equation for S.H.M. Simple harmonic motion. A particle is said to execute : A ‘ simple harmonic motion if it moves to and fro about a mean oe Suk Hence obtain expression for time position under the action of a restoring force which is directly proportional to its displacement from the mean ‘Differential equation of S.H.M. In SH.M,, the Position and is always directed towards the mean position, restoring force acting on the particle is proportional to If the displacement of the oscillating body from the _ its displacement. Thus mean position is small, then Fa-kx. Restoring force « Displacement ‘The negative sign shows that F and x are oppo- Fax or F=-kx sitely directed. Here k is spring factor or force constant.By Newton's second law, ee nt ar where 1 is the mass of the particle and acceleration @x ae ax ar or Put then te or fox dt’ (1) This is the differential equation of S.H.M. Here wis the angular frequency. Clearly, x should be such a function whose second derivative is equal to the function itself multiplied with a negative constant. Soa possible solution of equation (1) may be of the form x= Acos(wt +49) dx ‘The —=-oAsin (wt + en = @ASn(Ot + 6) and Fa -? Acos (ot + gy)=— ox 2 or fey otr=0 a which is same as equation (1). Hence the solution of the equation (1) is x= Acos(ot +5) 2) It gives displacement of the harmonic oscillator at any instant t. Here A is the amplitude of the oscillating particle. =o +, is the phase of the oscillating particle. dis the initial phase (at t =0) or epoch. Time period of S.H.M. If we replace t by t+ 2" in © equation (2), we get wl) A.cos (wl +2 + 49) = A cos (wt + 49) x 2a 2 ie, the motion repeats after time interval ~™. Hence —* © © Ee _ {inertia factor \ Spring factor is the time period of S.H.M. 22m, 2 or In general, nis called inertia factor and k the spring factor. OSCILLATIONS 14.5 14.8 SOME IMPORTANT TERMS CONNECTED WITH S.H.M. 9. Define the terms harmonic oscillator, displacement, amplitude, cycle, time period, frequency, angular frequency, phase and epoch with reference to oscillatory motion. ‘Some important terms connected with S.H.M. (i) Harmonic oscillator. A particle executing simple harmonic motion is called harmonic oscillator. (ii) Displacement. The distance of the oscillating particle from its mean position at any instant és called its displacement. It is denoted by x. ‘There can be other kind of displacement variables. ‘These can be voltage variations in time across a capacitor in an ac. circuit, pressure variations in time in the propa- gation of a sound wave, the changing electric and magnetic fields in the propagation of a light wave, etc. (iii) Amplitude. The maximum displacement of the oscillating particle on either side of its mean position is called its amplitude. Itis denoted by A. Thus X,,., =+ A (jv) Oscillation or cycle. One complete back and forth ‘motion of a particle starting and ending at the same point is called a cycle or oscillation o vibration. (®) Time period. The time taken by a particle to complete one oscillation is called its time period. Or, itis the smallest time interval after which the oscillatory motion repeats. It is denoted by T. (vi) Frequency. It is defined as the number of oscillations completed per unit time by a particle. Itis denoted by v (nu), Frequency is equal to the reciprocal of time period. That is, n T Clearly, the unit of frequency is(second)"' ors”. also expressed as cycles per second (cps) or hertz (Fz. SI unit of frequency = cps= Hz. (vii) Angular frequency. It is the quantity obtained by muiltiplying frequency v by a factor of 2x It is denoted bya Qn Thus, o=2av=2* SI unit of angular frequency = rad s“. (viii) Phase. The phase of a vibrating particle at any instant gives the state of the particle as regards its position ‘and the direction of motion at that instant. Itis equal to the argument of sine or cosine function occurring in the displacement equation of the S.H.M. Suppose a simple harmonic equation is represented by x= Acos(ot + dy) ‘Then phase of the particle is: 6= ot +14.6 _PHYSICS-XI : Clearly, the phase @ is a function of time f. It is usually expressed either as the fraction of the time period T or fraction of angle 2 that has elapsed since the vibrating particle last passed its mean position in the positive direction, $= ot + by 0 x= Acos (ot + ¢) -A ‘Thus the phase } gives an idea about the position and the direction of motion of the oscillating particle. (ix) Initial phase or epoch. The phase of a vibrating particle corresponding to time t =0 is called initial phase or epoch. Att=0, $=) The constant 6, is called initial phase or epoch. It tells about the initial state of motion of the vibrating particle 14.9 UNIFORM CIRCULAR MOTION AND S.H.M. 10. Show that simple harmonic motion may be regarded as the projection of uniform circular motion along a diameter of the circle. Hence derive an expre- ssion for the displacement of a particle in S. Relation between S.H.M. and uniform circular motion. As shown in Fig. 143, consider a particle P moving along a circle of radius A with uniform angular velocity w. Let N be the foot of the perpendicular drawn from the point P to the diameter XX". Then N is called the projection of P on the diameter XX’. As P moves along the circle from X to Y, Y to X', X’ to Y’ and Y' to X; N moves from X to O, Oto X’, X’ to Oand Oto X. Thus, as P revolves along the circumference of the Circle, N moves to and fro about the point O along the diameter XX’. The motion of N about O is said to be simple harmonic. Hence simple harmonic motion may be defined as the projection of uniform circular motion upon a diameter of a circle. The particle P is called reference particle or generating particle and the circle along which the particle P revolves is called circle of reference. Reference circle, Fig. 14.3 Displacement in simple harmonic motion. As shown in Fig. 14.4, consider a particle moving, in anticlockwise direction with uniform angular velocity along a circle of radius A and centre O. Suppose at time f =0, the reference particle is at point A such that ZXOA = by. At any time f, suppose the particle reaches the point P such that ZAOP = wt. Draw PN L XX’. Fig. 14.4 Displacement in S.H.M., epoch (+ ) Clearly, displacement of projection N from centre O at any instant fis x =ON. In right-angled AONP, ZPON = oot + by ON “op = cost +) cr X seos(ot+&) or x= Acos (at + 4). This equation gives displacement of a particle in S.HM. at any instant f. The quantity wt + @) is called phase of the particle and 4 is called initial phase or phase constant or epoch of the particle. The quantity A is called amplitude of the motion. It is a positive constant whose value depends on how the motion is initially started. Thus Phase ee x = A cos(at +b) f 1 t t Displacement Amplitude Angular Initial frequency phase Yom, x = Fig. 14.5 Epoch (-¢)As shown in Fig. 145, ifthe reference particle starts motion from the point P such that 2BOX =4, and ZBOP = ot, then ZPON = ol by x= Acos (ot ~ ty) Here ~ dyis the initial phase of the S.H.M. 11, Show that @ linear combination of sine and cosine functions like x(t) =asin ot +b cos wt represents a simple harmonic motion. Determine its ‘amplitude and phase constant. General expression for S.H.M. We are given x=asin ot +b cos at (1) Differentiating wart. time f, we get 8 — oa cos at - ob sin ot at Again, differentiating wart. time f, we get a <%= oFasin ot ~ a*b cos ot ae =- 0? (asin wt +b cos oot) ax 2 or oe e-wr ae ie, acceleration « displacement Hence the equation (1) represents S.H.M. To determine its amplitude and phase constant, we put Aces $ (2) and Asin (@) Then Acos $sin wt + Asin cos wt = A(sin at cos $+ cos wt sin 4) or Asin(ot +4) This again shows that equation (1) represents S.HM. of amplitude A and phase constant Squaring and adding (2) and (3), we get a? +b? = A? (cos? 6+ sin? $)= A? x1 © Amplitude, A =a? +b? Dividing (3) by (2), we get : b a Phase constant, ¢=tan-! © a 14.10 VELOCITY IN S.H.M. 12. Deduce an expression for the velocity of a particle executing S.H.M. When is the particle velocity (i) maximum and (ii) minimum 2 OSCILLATIONS 14.7 Expression for the velocity of a particle executing S.H.M. As shown in Fig. 14.6, consider a particle P moving with uniform angular speed « in a circle of radius A. Its velocity vector v is directed along the tangent and the magnitude of this velocity vector is v= Angular velocity x radius =A Fig. 14.6 Velocity of a particle in S.H.M. Draw PP’ and QQ perpendiculars to the diameter XX’. The motion of P'is simple harmonic. Clearly, the instantaneous velocity of a particle executing S.H.M. will be v(1)= Velocity of the particle P’ at any instant f = Projection of the velocity » of the reference particle P = PQ =PQ=-vsin(ot + 4) or v(t)=-@ Asin (ot + dy). ‘The negative sign shows that the velocity of P is directed towards left i,, in the negative X-direction. Moreover, a z 0(t)=-@A 1 cos" (wot + $9) =- 0A 1 a or v(t)=-w A? -x? Special cases. (i) When the particle is at the mean position, then x =0, so v(t)=-@ A? -0? =-0A This is the maximum velocity which a particle in SLM. can execute and is called velocity amplitude, denoted by Pray [ex =A cos (ot + O91 =0A Pmax (ii) When the particle is at the extreme position, then x= A 50 v=-oJA?- A? Thus the velocity of a particle in SHM. is zero at either of its extreme positions.14.8 PHYSICS-XI ACCELERATION OF A PARTICLE IN S.H.M. 13. Show that the acceleration of a particle in S.H.M. is proportional to its displacement from the mean position. Hence write the expression for the time period of S.H.M. Expression for the acceleration of a particle executing S.H.M. As shown in Fig. 14.7, consider a particle P| moving with uniform angular speed q in a circle of radius A. The particle has the centripetal acceleration acting radially towards the centre O, The magnitude of this acceleration is a, = 7A 14.11 Fig. 14.7. Acceleration of a particle in S.H.M Draw PP and QQ’ perpendiculars to the diameter XX", The motion of P’ is simple harmonic. Clearly, the instantaneous acceleration of a particle executing SHM. will be a(t) = Acceleration of particle Pat any instant ¢ = Projection of the acceleration a, of the reference particle P = Projection of PQ on diameter XX’ = PQ =a, cos (w! + 49) or a(t)=~a? A cos (wt + §)) =- are This equation expresses the acceleration of a particle executing S.H.M. It shows that the acceleration of a particle in S.H.M. is proportional to its displacement from the mean position and acts in the opposite direction of the displacement. Special cases. (i) When the particle is at the mean position, then x=0, so, acceleration =- w? (0) =0. Hence the acceleration of a particle in S.H.M. is zero at the mean position. (ii) When the particle is at the extreme position, then x= Aso, acceleration =— oA This is the maximum value of acceleration which a particle in SH.M. can possess and is called acceleration amplitude, denoted by 4,4, ol 2 14.12 PHASE RELATIONSHIP BETWEEN DISPLACEMENT, VELOCITY AND ACCELERATION 14. Draw displacement-time, velocity-time and acceleration-time graphs for a particle executing simple harmonic motion. Discuss their phase relationship, Inter-relationship between particle displacement, velocity and acceleration in S.H.M. If a particle ‘executing S.H.M. passes through its positive extreme Position (x =+ A) at time ¢ =0, then its displacement equation can be written as x()= Acos at Asin ot Velocity, atts pa o cos( ) oF A cos ot Acceleration, a(t)=@ a = @ A cos (wt + 7). Using the above relations, we determine the values of displacement, velocity and acceleration at various instant ¢ for one complete cycle as illustrated below. Time, ¢ o|e ltl] + Praseangleat= 281] 9 | = |, Ele Displacement. x(t) | +a] 0 | -a] 0} va veosyoy | 0 [wr] 0 [ead] 0° ‘ecicatona@) [=a] 0 frwta] 0. [mata In Fig. 14.8, we have plotted separately the x versus t, » versus f and a versus t curves for a simple harmonic motion. | +4 | Displacement —> Velocity + Acceleration + Fig. 14.8 Relation between velocity, displacement and acceleration in S.H.M.OSCILLATIONS 14.9 Conclusions. From the above graphs, we ean draw the following conclusions about simple harmonic motion : () Displacement, velocity and acceleration, all vary harmonically with time. ExAMur 3. The following figures depict two circular ‘motions. The radius ofthe circle, the period of revolution, the initial position and the sense of revolution are indicated on the figures, Obtain the simple harmonic motions of the x-projection of the radius vector of the rotating particle P in each case, INCERTI (i) The velocity amplitude is o times ; and acceleration amplitude is o* times the displacement amplitude A, (iii) Clearly, the velocity curve lies shifted to the left of the displacement curve by an interval of T/4. Thus. the particle velocity is ahead ofits displacement by a phase angle of x/2 rad. This means that whichever value displacement attains at any instant, velocity attains a imilar value a T/4 time (a quarter of cycle) earlier. When the particle velocity is maximum, the displace- ment is minimum and vice versa. (iv) Clearly, the acceleration curve lies shifted to the left of the displacement curve by an interval of T/2. Thus the particle acceleration is ahead of its displacement by @ phase angle of x rad. Or, acceleration is ahead of velocity in phase by x/2 rad. When acceleration has ‘maximum positive value, displacement has maximum negative value and vice versa. When the displacement is zero, the acceleration is also zero. Ypt=0) Solution. (a) As shown in Fig, 14.10(a), suppose the particle moves in the anticlockwise sense from P to P in time t. Angle swept by the radius vector, 28 ce ‘Nis the foot of perpendicular drawn from P’on the axis. 0=ot =2h, fe T=45] Formutar Usto 1. Displacement, = Acos (at + ) where A= amplitude, w= angular frequency and 4 = initial phase of particle in SHM. Velocity, v= £ @ Asin (at + 4) gattiion, =-0/4* ‘Acceleration, Maximum acceleration, 4, Restoring force, F = ~ kx where k = force constant and w? =_k/m 5. Angular frequency, w= 2x v = 2n/T. (Displacement Ti t ae as el ny Kesler (inertia factor 7. Time period, T= 2 = aaa Spring factor Units Useo, Displacement x and amplitude A are in m or em, force constant k in Nm, frequency v in Hz, angular frequency w in rads“). Displacement, ON = OP cos (8 + x/4) 22,4 4 i) This represents S.H.M. of amplitude a, period 4 s and an initial phase = r/4 rad. or ¥(= 4 cos( Fig. 14.10 (®) As shown in Fig. 14.10(b), suppose the particle moves in the clockwise sense from P to Pin time f. Angle swept by the radius vector,14.10 PHYSICS-xI Displacement, ON = OP eo or (= eos _ or x (y= boos (3 This represents S.H.M. of amplitude b, period 30 s and an initial phase =~ x/2 rad. ENAwPut: ge A simple harmonic motion is represented by x =10 sin (20 t +05) Write down its amplitude, angular frequency, frequency, time period and initial phase, if displacement is measured in metres and time in seconds. (Himachal 09C} Solution. Given x =10 sin (20 t +0.5) Standard equation for displacement in SHM is Asin (ot + 4) ‘Comparing the above two equations, we get () Amplitude, A=10 m. [y Aand x have same inits} (ii) Angular frequency, = 20 rad 5". 2 2 Ri Wy (ii) requency, v= => (iv) Time period, T (®) Initial phase, 4, = 0.5 rad. Exampre 5. A particle executes SHM with a time period of 2s and amplitude 5 cm Find (i) displacement (ii) velocity and (iii) acceleration, after 1/3 second ; starting from the mean position. Solution, Here 3, A=Sam, t=1/3s (@ For the particle starting from mean position, Displacement, x= A’sin ot =A sin 2 t (ii) Velocity, v _2nx5 zi cos F=5x 3.1405 =7.85 em xammple 6. A body oscillates with SHM according to the equation X(1) =5c0s (2at-+ x/ 4), where t is in see. and x in metres. Calculate (a) Displacement at t (b) Time period (0) Initial velocity [Central Schools 08] Solution. Given x(1)=5 cos(2nt + n/4) We compare with standard equation, x(t) = Acos(iot + 6) (a) Displacement at t =0, Ie 200) =Scos =5x =m (b) Clearly, w=2n or 7 On Time period, T=15. dx Velocity, v= (6) Velocity, v= Ssin(2at + Phan Initial velocity at #=0, since Examen: 7. A body oscillates with SHM according to the ‘equation, x =(5.0 mi) cos [(2x rad s"') t+ 1/4} At t=15 55, calculate (a) displacement, (b) speed and (C) acceleration of the body. + INCERT} Solution. Here =2n rads, T=2n/o=15, tH15s (@) Displacement, O cos (22x15 + x/4) (b) Velocity, dx _d So fp 150 05 nt + /4)] 5.0x2nsin Qt + n/4) 1.0 x 2 sin (2nx 1.5 + x/4) =+50%2nsin x/4=50%2x 2 x0707 = 22.22 ms (c) Acceleration, SZ t-tonsin et + x/4)) 20m? cos (2nt + n/ 4) 4n? [5.0 cos (2nx 1.5 + x/4)] 4x 9.87 x (~3.535) = 139.56 ms", [Using (a)]apne: 8. The equation of a simple harmonic motion is ‘given by y =6 sin 10x ¢ +8 cos 10x t, where y is in cm and 1 in sec. Determine the amplitude, period and initial phase. Solution. Given y =6 sin 10x +8 cos0xt (1) The general equation of SHM is y= asin (of + @)=asin of cos +a cos at sin } =(a.cos 4) sin wt + (a sin ) cos «ot 2) ‘Comparing equations (1) and (2), we get acos $=6 (3) asin o=8 A) and ot=10nt or 2a _2n Time period, T ime pe Tr o Squaring and adding (3) and (4), we get 2 yg? =36 +64=100 or a” =100 <. Amplitude, a= 10 cm. Dividing (4) by (3), we get tan > 1.3333, a (cos? $+ sin? §) . Initial phase, = tan” ' (1.3333) = 53°8'. Exaapr 9. A particle executes S.H.M. of amplitude 25 cm and time period 3 s. What is the minimum time required for the particle to move between two points 12.5 am on either side of the mean position ? Solution. When the particle starts from mean position, its displacement at instant t is given by y=Asinot Given A=25em, T=35, y=125cm or 2: Time taken by the particle to move between two points 12.5 cm on either side of mean position is given by 5s. 4 Exaniput 10. The shortest distance travelled by a particle executing SHM from mean position in2 sis equal to(/3 /2) times its amplitude. Determine its time period. Solution. Here t=2s, y=(v3/2)A, T As 2 = Asin wt = Asin t y= Asin ot = Asin = OSCILLATIONS 14.14 or T=12s. anna: 11. The time-period of a simple pendulum is 2 5 or ‘and it can go to and fro from equilibrium position at a maximum distance of 5 cm Ifat the start of the motion the pendulum is in the position of maximum displacement towards the right of the equilibrium position, then write the displacement equation of the pendulum. Solution. The displacement in SHM is given by y= Asin (wt + 65) Given T=2s, A=5em 2m 28 ered st 2 y=5ssin (nt + dy) | displacement y =5 cm. Therefore, 5=5sin(mx0+6) sin )=1 qan2 Hence displacement equation for the pendulum is Attime t or =Ssin( rts £) =5.cosnt 2 EXAMPLE 12. A particle executes $.H.M. of time period 10 seconds. The displacement of particle at any instant is given by : x =10 sin ct (in cm). Find (i) the velocity of body 2safter it passes through mean position (i) the acceleration 2s after it passes the mean position. (Central Schools 041 Solution. Here T=10s, x=10 sin wt em dx 1 (i) Velocity, 9 =F =10 «cos at em 5 -10(2 Joo tom tT) T Velocity of the body 2s after it passes through the mean position, v=10(22)cos(2#x2)em 5° 10) * (a0 = 2m cos 72° =2 x 3.14 0.309 = 1.94 ems. (ii) Acceleration of the body 2s after it passes through the mean position, = 4*987%0.951 76 omg 1014.12 PHYSICS-XI E:Aweus 13. For a particle in SHM, the displacement x of the particle as a function of time tis given as x= Asin(2 xt) Here x is in cm and t is in seconds. Let the time taken by the particle to travel fron x =0 to x= A/2 be t, and the time taken to travel fiom x= A/2 to x= Abe ty. Find ty [ty {Delhi 04) Solution, Here 2n Also o or th= >=98 or Exasurts 14. Ina HCI molecule, we may treat Cl to be of infinite mass and H alone oscillating. If the oscillation of LCT molecule shows frequency9 x 10" s-1, deduce the force sant, The Avogadro number =6 x 10° per kg-mole. Solution. Frequency, v=9x 10" s-! Mass of a H-atom, m= = 533.4 Nm“, ExanrLé 1g. A particle is moving with SHM in a straight livie. When the distance of the particle from the equilibrium position has values x, and x,, the corresponding values of velocities are u, and us, Show that the time period of oscillation is given by 7 T=2n . Solution. When xaXy, When x=x, v= or Pa k(a? (1) and z = 0A ~(2) Subtracting (2) from (1), we get 1B 1B = @? (A? 32)? (A* 18) =o (x3 - 23) 1 Baud a oat Examrit: 16. If the distance y of a point moving on a straight line measured from a fixed origin on it and velocity vare connected by the relation 4v =25 —y?, then show that the motion is simple harmonic and find its time period. Solution. Given 4v? =25 ~ 7 ey Also velocity in SHM, v= JA? —y? Comparing the above two equations, we find that the given equation represents SHM of amplitude A =5 and @=1/2 rads Time-period, T or Qn _2nx2 o 1 EXAMPLe 17. A particle executing SHM along a straight line has a velocity of 4 ms? when at a distance3 mfrom the ‘mean position and 3 ms~' when at a distance of 4 mifrom it Find the time it takes to travel 2.5 mfrom the positive extre- rity of its oscillation. Solution. When y, =3 m, =4ns. As or (1) and 3=@y A? ~4? or9 =a? (A216) ..(2) Dividing (1) by (2), we get oe or 16 A? ~256 =94? -81 9 A W16 or 7 A? =256 -81=175 or A? =25 : A=VB=5m From (1), 4= 05-3? or rads?When the particle is 25 m from the positive extreme position, its displacement from the mean position is y=5-25=25m When the time is noted from the extreme position, we can write y= Acos wt 25-5 cos(Ix t) este or 32.3 Hence t= ¥=2242 - 1.0475. a) A particle executing linear SHM has a ty of 40 ems"! and a maximum accele- ration of 50. ons”. Find its amplitude and the period of oscillation. Solution. Maximum velocity, Vmax = A= 40 em st Maximum acceleration, or 40x4 Amplitude, A=" =. =32cm, Period of oscillation, pa2B 2x31 x4 ogo @ 5 Exar 19. The vertical motion of a huge piston in a ‘machine is approximately simple harmonic with a frequency of 0.50 s"'. A block of 10 kg is placed on the pistort. What is the maximum amplitude of the piston’s SHM for the block ‘and the piston to remain together ? : Solution. Here v=05s"!, g=9.8ms~ ‘The maximum acceleration in SHM is given by nay =F A=(Qnvl A=4n?v? A ‘The block will remain in contact with the piston if Ayax $8 oF An VE ASR Hence the maximum amplitude of the piston will be g r Anox = 722 “GOS? anv Examput 20. A block of mass one kg is fastened to a spring ‘with a spring constant 50 Nui” !. The block is pulled to a OSCILLATIONS 14.13 distance x =10.cm from its equilibrium position at x =Oon a {frictionless surface from rest at t =0. Write the expression for its x(t) and v(t). {Central Schools 03] Solution. Here m=1kg, k=50Nm7', A =10 em =0.10 m. f = =7.07 rad 6". m - ‘As the motion starts from the mean position, so the displacement equation can be written as or x(#)=0.10 sin 7.07 ¢ o x(t)= Asin of and 2(t)= & =010%707 ¢0s7.07 t or v(t)= 0.707 cos 7.07 ms”. Exanrot 2, A person normally weighing 60 kg stands on a platform tohtch oscillates up and down harmonically at @ frequency of 2.0 5-' and an amplitude 5.0 cm If a machine ‘on the platform gives the person's weight against time, deduce the maximum and minimum readings if will show. Take g =10 ms~*. Solution. The platform vibrates between the posi- a Platform tions A and B about the fs mean position O, as shown in Fig. 14.11. mg Given A=5.0 cm, ‘ oem 40 m=60 kg v=2 Hz ' At Aand B the accele- ; ration is maximum and is directed towards the mean ry position. B Itis given by mg 2 a oA Fig. 14.11 nv? A = 4x 9.87 x (2)? x 0.05 =7.9 ms At A both the weight mg and the restoring force F are directed towards O. Therefore, the weight at A is maximum and is given by W, =(mg + F)=(mg + maya) = P18 + Mypax) 60 (10 + 7.9) =60 x 17.9 =1074 N 1074 _ 1074 ¢ 1 At B mg and F are opposed to each other so that the weight is minimum. It is given by Wy = (mg ~ F) = (img ~ myyy,)= (8 ~ Menax) = 60 (10 -7.9) =(60 x 2.1) N =126N 126 aap 2 ks f 107.4 kg f.14.14 PHYSICS-XI Lane 22. A body of mass 0.1 kg is executing SHM ‘according to the equation a 205 os( 1001 +2) metre : ( 4 Find (i) the frequency of oscillation (i) initial phase (ii) maximum velocity (iv) maxintum acceleration and (v) total energy. Solution. Given x =0.5 cos{ 100 1+ %) metre For any SHM, x= Acos(t + 4) ‘Comparing the above two equations, we get A=05m, @=100 rads, ao rad (i) Frequency, v= z (ii) Initial phase, 4, (iii) By, = @ A=100 x 0.5 = 50 ms“, (2) a. = 0" A=(100) «0.5 = 5000 ms, (2) Total energy =2 m2, = 2M max 3801 x (50)? =125 J. ® PROBLEMS For Practice 1. Assimple harmonic oscillation is represented by the equation, y= 0.40'sin (4401 + 0.61) Here y and t are in m and s respectively. What are the values of () amplitude (ii) angular frequency (iii) frequency of oscillations (iv) time period of oscillations and (0) initial phase ? [Ans. (i) 0.40 m (fi) 440 rad s™ (ji) 70 Hz (iv) 0.0143 s (v) 0.61 rad] 2. The periodic time of a body executing SHM is 2 s. After how much time interval from ¢ = 0, will its displacement be half of its amplitude ? (Ans. 1/65) 3. A particle executes SHM represented by the ‘equation : 10y = 0.1 sin 50 mt, where the displacement y is in metre and time ¢ in second. Find the amplitude and frequency of the particle. (Ans, A =0.01m, v= 25 Hz) 4. The displacement of a particle executing periodic motion is given by y = 4cos* (t /2) sin (1000 t) Find the independent constituent SHM’s. IMT 93] [Ans. sin (10011) sin (1000#) sin (999¢)} 5. A particle executing SHM completes 1200 oscilla- tions per minute and passes through the mean Position with a velocity of 31.4 ms~'. Determine the 10. 11. 2 1B. 14. 15. maximum displacement of the particle from the mean position, Also obtain the displacement equation of the particle if its displacement be zero at the instant ¢ [Ans, A=0.025 m, y= .025 sin (40nt) metre] The acceleration of a particle performing SHM is 12cm s* at a distance of 3 em from the mean position. Calculate its time-period. (Ans. 3.142 s) In a pendulum, the amplitude is 0.05 m and a period of 2 s, Compute the maximum velocity. (Ans. 0.1571 ms“) In what time after its motion begins, will a particle oscillating according to the equation, y= 7sin 0.5 x, move from the mean position to maximum displacement ? [Himachal 08C] (Ans. 1s) A particle executes SHM on a straight line path. The amplitude of oscillation is 2m. When the displace- ment of the particle from the mean position is 1 cm, the magnitude of its acceleration is equal to its velocity. Find the time period, maximum velocity and maximum acceleration of SHM. (Ans. 3.635, 3.464 ems”, 6cms” 2) The velocity of a particle describing SHM is ‘16cm s“ at a distance of 8 cm from mean position and 8cms” at a distance of 12cm from mean Position. Calculate the amplitude of the motion. (Ans. 13.06 cm) A particle is executing SHM. If 4, and u, are the speeds of the particle at distances x, and x, from the equilibrium position, show that the frequency of oscillation, 4 1/2 nel4) - If a particle executes SHM of time period 4 s and amplitude 2 cm, find its maximum velocity and that at half its full displacement. Also find the acceleration at the turing points and when the displacement is 0.75 em. (Ans. 3.14em 5“, 272cms"', 493em s*, 1.85¢m $4) Show that if a particle is moving in SHM, its velo- city at a distance V3 / 2 of its amplitude from the cen- tral position is half its velocity in central position. [Chandigarh 03 ; Central Schools 09] A particle executes SHM of period 12 s. Two seconds after it passes through the centre of oscillation, the velocity is found to be 3.142cm s™. Find the amplitude and the length of the path. (Ans. 12 cm, 24 cm) A block lying on a horizontal table executes SHM of period 1 second, horizontally. What is the maximumamplitude for which the block does not slide ? Coefficient of friction between block and surface is, 04 x? =10. (Ans. 9.8 cm) 16. A horizontal platform moves up and down simple harmonically, the total vertical movement being, 10 cm. What is the shortest period permissible, if objects resting on the platform are to remain in contact with it throughout the motion ? Take 3 =980cm 5, (Ans. 0.449 s) 17. In a gasoline engine, the motion of the piston is simple harmonic. The piston has a mass of 2 kg and stroke (twice the amplitude) of 10 cm. Find maxi- mum acceleration and the maximum unbalanced force on the piston, if it is making 50 complete vibrations each minute. (Ans. 1.371 ms", 2.742 N) 18. A man stands on a weighing machine placed on a horizontal platform. The machine reads 50 kg. By ‘means of a suitable mechanism the platform is made to execute harmonic vibrations up and down with a frequency of 2 vibrations per second. What will be the effect on the reading of the weighing, machine ? The amplitude of vibration of the platform is 5 cm. Take g = 10 ms. (Ans, Max. reading = 895 kg f, Min. reading = 10.5 kg ®) x HINTS 1. Comparing y = 0.40sin (440 + 0.61) with y= Asin (ot +49), we get ( Amplitude, A= 0.40 m. (ii) Angular frequency, (iii) Frequency, (iv) Time period, (2) Initial phase, 9 = 0.61 rad. 2 Here T=2s, y=A/2 t=? As y= Asin of = Asin “Te t=1/68. 4. ¥=4cos" (t /2)sin (1000¢) =2(1+ cos t)sin (1000) [1+ cos 20 = 2cos? @ =2sin (1000) + 2sin (1000) cos t =2sin (1000) + [sin (1000¢ +t) + sin (1000¢ -1)] [+ 2sin Acos B=sin (A+ B)+sin(A~ B)) =2sin (1000t) + sin (1001) + sin (999) 10. OSCILLATIONS 14.15 ‘Thus motion y is composed of three independent SHMs which are sin (1000#),sin (1001) and sin (9991), Here y= 1200 0 Upae =O A=2nv A 314 x3, 2Hz, y= 314s! But Az tmx = anv = 0.025 m. x20 As displacement is zero att = 0, so we can write y= Asin ot = Asin (2n vt) = 0.025 sin (40nt), Herea=12ems, y=3em on figfBeainae® y Time period, T=2% =?2*3.1 _ 5149 2n mae WARE A y Ast = 3.142 0.05 Given 1571 ms". y=7sin 05 xt On comparing with y=asin ot, we get the standard equation, 2=7,0=05K Let f be the time taken by the particle in moving from mean position to maximum displacement. Then 7=7sinOSnt orsinOSxt= . OSnt== or 2 Here A=2cm. When displacement y = 1cm, magnitude of velocity = magnitude of acceleration or wfA?=P=aky « Rapowy or 2-PeoxP or w=V3 rads :. Time period, TF 3.695, Umax = @ A = V3 x2 = 1.732 x2=3.464em 7, ms? As eae Infirstease: 16-0 {=H In second ase :8= 0 (AP =TF Dividing (1) by (2), of |, [ea 8 afew 7 Var On solving 3A*=512 or A?=1706 or A=13.06 em.14.16 PHYSICS-XI 12. Here T=4s, A=2em, 2m 2x34 1 57 rads! tT # yyy = A= 157 «22 3.14em ot, At /2=1em, 97 At the turning points, acceleration is maximum ? A= (1.57) x2= 4.93 em $7. 272m veaya— v Anas = At y=0.75em, a= 07 y=(157)* «0.75 = 1.85 em 5. 13, Here y=V3/2A eo dae veo A = joas 14 Let Then 3.14% or Ba x98 x(1P 28%(" _ 0098 m = 9.8 cm, 4x10 2 an) 1. Take any, =0? A=(2*) A= 16 ke tagcat Aa 2) Ae 17 Length of stroke = 2 A = 10cm. 18, Here m=50kg, v=2Hz, A=Sem =005 m = 4987 x2 «005 a, ev? A Max. force on the man =I (§ + yyy = 50(10 + 7.9)= 8950 N = 89.5 kg f. Min, force on the man =m (§ ~ Gy.) = 50(10 - 7.9) = 105.0 N =105 kg f. 14.13 ENERGY IN S.H.M. : KINETIC AND POTENTIAL ENERGIES 15. Derive expressions for the kinetic and potential energies of a simple harmonic oscillator. Hence show that the total energy is conserved in S.H.M. In which positions of the oscillator, is the energy wholly kinetic or wholly potent al ? Total energy in S.H.M. The energy of a harmonic oscillator is partly kinetic and partly potential. When a body is displaced from its equilibrium position by doing work upon it, it acquires potential energy. When the body is released, it begins to move back with a velocity, thus acquiring kinetic energy. (i) Kinetic energy. At any instagt, the displacement of a particle executing S.H.M. is given by x= Acos(ot +4) dx Velocity, v= elocity, v= Hence kinetic energy of the particle at any displacement x is given by 2 =o Asin (ot + 4) 1 2 A? sin? K=— mv’ mor A® sin? (cot + a (ot +4) 2 But A? sin? (of + 4) = A* [1 —cos? (ot + 4)] = A? — A? cos" (cot + 4) = A? 2? K 1a? at +4) Ka mat (4? —32)=2 (4? 24) or (ii) Potential energy. When the displacement of a particle from its equilibrium position is x, the restoring force acting on it is Fa-kx If we displace the particle further through a small distance dx, then work done against the restoring force is given by dW =-Fdr=+kxdx The total work done in moving the particle from mean position (x =0) to displacement x is given by x 2 W=f w= f kxdr=k|= ; 2 This work done against the restoring force is stored as the potential energy of the particle. Hlence potential energy of a particle at displacement x is given by =)? ° 2 ma? A? cos* (wt + d) (ii) Total energy. At any displacement x, the total energy of a harmonic oscillatory is given by 1 B= k+u=2e(a?-24) 4165 und e(at ays Die o OE grata} PAP =2n? mv? AP : [:@=2m) Thus the total mechanical energy of a harmonic oscillator is independent of time or displacement. Hence in the absence of any frictional force, the total energy of @ harmonic oscillator is conserved.Obviously, the total energy of particle in S.H.M. is (i) directly proportional to the mass of the particle, i) directly proportional to the square of its frequency v, and (iii) directly proportional to the square of its vibra- tional amplitude A. Graphical representation. At the mean position, x =0 Kinetic energy, K=4k(A®-02)=4 kA? Potential energy, U=3* (0*)=0 Hence at the mean position, the energy is all kinetic, ‘At the extreme positions, x= A Kinetic energy, K=4k (A? - A?) =0 Potential energy, U=4 ka? Hence at the two extreme positions, the energy is all potential. Figure 14.12 shows the variations of kinetic energy K, potential energy Land total energy E with displace- ment x. The graphs for K and U are parabolic while that for Eis a straight line parallel to the displacement axis. At x =0, the energy is all kinetic and for x=+ A the energy is all potential. Energy E=k+U oO x=+A + Displacement —- Fig. 14.12 K, U and E as functions of displacement x for a harmonic oscillator. Figure 14.13 shows the variations of energies K, U and E ofa harmonic oscillator with time f. Clearly, twice in each cycle, both kinetic and potential energies assume their peak values. Both of these energies are periodic functions of time, the time period of each being T/2. Energy oT 17 ama oT Time (1) Fig. 14.13 K, U and E as functions of time t for a harmonic oscillator. OSCILLATIONS 14.17 Examples based on Formutae UseD 1. PE. at displacement y from the mean position, 1 1 digo mat Piet Lg? 2} mo? Z =4 mo? A? sin? ot ghhagme ena 2. KE. at displacement y from the mean position, RAPP) =Z mo? (A =P) 4s a? abcca® =} mo? A? cos? ot 2 3. Total energy at any point, mo? A? 22x? m ABV? Unis Useo Energies E,, E, and Eare injoule, displacement in metre, force constant k in Nm“! and angular frequency o in rads“ Exanrut: 23. A block whose mass is 1 kg is fastened to a Spring. The spring has a spring constant of 50 N ni "The block is pulled to a distance x =10 cm from its equilibrium position at x =0 on a frictionless surface from rest at t =0. Calculate the kinetic, potential and total energies of the block when it is 5 cm away from the mean position. (NCERT] Solution. Here m=1kg, k=50N mt, A=10 cm =0.10 m, y=5 cm =0.05 m Kinetic energy, E=2k(A #)=4x 50[0.107 —(0.05)"} = 0.1875 J. Potential energy, E,= hy? =4x 50 x (0.05)? = 0.0625 J. Total energy, E= E, +E, =0.1875 + 0.0625 = 0.25 J. Exam A body executes SHM of time period 8 If its mass be0-1 kg its velocity 1 second after it passes through its mean position be 4 ms~', find its (j) kinetic energy (ii) potential energy and (iii) total energy. Solution. Here m=O.1kg, T=85 When But or14.18 PHYSICS-xi Total energy, mmo? a=} coax (S) x 16 2 4 ® 6 J. Kinetic energy, 1 oax(4y= 7 x01x (4) =08 J. Potential energy, E, = E-E, =1.6-08=08 J. spring of force constant 800 Nm" has an extension of 5 cm. What is the work done in increasing the extension from 5 om to 15 am? IATEEE 02] Solution. Here k =800 Nm~', x, =5 cm =0.05 m, 15cm =0.15 m W = Increase in P-E. of the spring Examen =1ku2 =4k(d = }* 800 [(0.15)? - 0.057] J 8. A particle of mass 10 g is describing SHM along a straight line with a period of 2 s and amplitude of 10 cm What is the kinetic energy when itis (i)2 em(ii) 5 om ‘rom its equilibrium position ? How do you account for the difference between its two values ? Solution. Velocity at displacement y is ia v=0fA-y¥ Given A=10 cm, T Angular frequency, 2e eT crate wo () When y=2 em, v= 100-4 = 2 96 cms KE=} mo? F105 12 «96 = 480 x? erg. (i) When y =5 cm, 100-25 =n V75 cms 7% 10x 1x75 =375 x? erg. The K.E. decreases when the particle moves from y=2 cm to y=5 cm. This is due to the increase in the potential energy of the particle. bxawri Ata time when the displacement is half the ‘amplitude, what fraction of the total energy is kinetic and what fraction is potential in S.H.M. ? Solution. Displacement 3 amplitude ory Total energy of SHM, E 3 moka? Kinetic energy of SHM, E, = 3 mo? (A? ~ y2 E, 1 Lala 42 EXAMPLE 28. A particle is executing SHM of amplitude A. At what displacement from the mean position, is the energy half kinetic and half potential? Solution. As, x or or ‘Thus the energy will be half kinetic and half potential at displacement + on either side of the mean position. Exams: 20. A particle executes simple harmonic motion ‘of amplitude A. (i) At what distance from the mean position is its kinetic energy equal to its potential energy ? (ii) At ‘what points is its speed half the maximum speed ? Solution. The potential energy and kinetic energy of a particle at a displacement y are given by and (0) where A is the amplitude and kis the force constant.771 times the amplitude on either side of mean position. i me? : st ix Maximum kinetic energy 1 B, =4x(E, Q) cs $x (Emo @) From equation (1), 1 B=ik 2 ( te frat [Put y =0} Putting these values in equation (2), we get Lea? tue lel nat daa - xt ka’ pe cag 2 or 4y? =34? a or eS aq2036 8 = 0.86 times the amplitude on either side of mean position. X% PROBLEMS FoR PRacTICE 1. Abob of simple pendulum of mass 1 g is oscillating with a frequency 5 vibrations per second and its amplitude is 3 cm. Find the kinetic energy of the bob in the lowest position. (Ans. 4441.5 erg) 2. A body weighing 10 g has a velocity of 6cm s™ after one second of its starting from mean position, If the time period is 6 seconds, find the kinetic energy, potential energy and the total energy. (Ans. 180 erg, 540 erg, 720 erg) 3. A particle executes SHM of period 8 seconds. After what time of its passing through the mean position will the energy be half kinetic and half potential ? [Chandigarh 08] (Ans. 1 5) OSCILLATIONS 14.19 4, The total energy of a particle executing SHM of period 2n seconds is 1.024 x 10° J. The displace- ment of the particle at x/ 4s is 0.08V2 m. Calculate the amplitude of motion and mass of the particle. (Ans. 0.16 m ; 0.08 kg) 5. A particle which is attached to a spring oscillates horizontally with simple harmonic motion with a frequency of 1/ x Hz and total energy of 10 J. If the ‘maximum speed of the particle is 0.4 ms", what is, the force constant of the spring ? What will be the maximum potential energy of the spring during the (Ans. k= 500 Nm", Ups, = 103) 6. The length of a weightless spring increases by 2cm when a weight of 1.0 kg is suspended from it. The weight is pulled down by 10 cm and released Determine the period of oscillation of the spring and its kinetic energy of oscillation. Take g = 10ms~*. (Ans, 0.28 s, 2.5 J) motion ? X HINTS 1. At the lowest or the mean position, energy of the bob is entirely kinetic and maximum. aa E ==ma A’ (m5 2. Herem=10g, T=6s, ee 2 ast o-% rad Ts When t=1s, v= 6cms" As v=Aecos ot Kinetic energy . Potential energy = Total energy — Kinetic energy 720 - 180 = 540 erg.14.20 PHYSICS-xI Now 2n = Asin ot = Asin 1 y= Asin sin 0.08 V2 =a sin 2 or 008 V3 = Ax A= 0.08 V2 x V2 = 0.16 m. : 2s Tost energy Lina? Inf 28) 31), (2e 1.024 10-3 =2 2)? (a6) one ( io 21.024 «10°? “(a6 5. Here v=1/nHz, E=10J, v,,,=04ms"* or om = 0.08 kg. Now t=O A=2nv A Unay _O4XR Aq tna Qnv Qexd As Exdkat 2 2E 2x10 (027 =10). 6. Here F=mg=1.0x10N, y= F_10%10 kat vy 002 im 10 =2x314x k 500 , = Work done in pulling the spring through. 10cm or 0.1m =500 Nm“ (Ep)max = fem = 0.02 m 00 Nm" T=2n .28 5. Vat 2 =tkx? =] <500x(0.)? = 3 kx? = 500 x(0.17 = 2.5 J. 14.14 OSCILLATIONS DUE TO A SPRING 16. Derive an expression for the time-period of the horizontal oscillations of a massless loaded spring. Horizontal oscillations of a body on a spring. Consider a massless spring lying on a frictionless horizontal table. Its one end is attached to a rigid support and the other end to a body of mass m If the body is pulled towards right through a small distance x and released, it starts oscillating back and forth about its equilibrium position under the action of the restoring force of elasticity, Fank where k is the force constant (restoring force per unit ‘compression or extension) of the spring. The negative sign indicates that the force is directed oppositely to x. Equilibrium [“DOOTTUTHT IOUT] ms Compressed bes Fig. 14.14 Horizontal oscillations of a loaded spring. If d?x/ dt? is the acceleration of the body, then it. ar P, ae im This shows that the acceleration is proportional to displacement x and acts opposite to it. Hence the body executes simple harmonic motion. Its time period is given by ra28_2e @ vkim or Ta2n ft k Frequency of oscillation leek veda Clearly, the time period T will be small or frequency v large if the spring is stiff (high k) and attached body is light (small m). 17. Deduce an expression for the time-period of the vertical oscillations of a massless loaded spring. Does it depend on acceleration due to gravity ? Vertical oscillations of a body on a spring. If a spring is suspended vertically from a rigid support and a body of mass m is attached to its lower end, thespring gets stretched to a distance d due to the weight ig of the body. Because of the elasticity of the spring, restoring force equal to kd begins to act in the upward direction. Here k is the spring factor of the spring. In the position of equilibrium, mg =kd If the body be pulled vertically downwards through a small distance x from its equilibrium position and then released, it begins to oscillate up and down about this position. The weight mg of the body. acts vertically downwards while the restoring force Kk(d+x) due to elasticity acts vertically upwards. Therefore, the resultant force on the body is F=mg-k(d+x) kd —kd ~kx ke [emg = kd or Equilibrium Stretched Equilibrium Displaced position Fig. 14.15 =-wr Thus acceleration is proportional to displacement x and is directed opposite to it. Hence the body executes S.H.M. and its time period is Qn Qn mt 2k TE raze, o ykin & Obviously, the force of gravity has no effect on the force constant kand hence the time period of the oscillating mass. 14.15 OSCILLATIONS OF LOADED SPRING COMBINATION 18. Fig. 14.16 shows four different spring arrange- ments. If the mass of each arrangement is displaced from its equilibrium position and released, what is the resulting frequency of vibration in each case ? Neglect the mass of each spring. INCERT] OSCILLATIONS 14.21 h mi Ky | OOOTTTOTT | s : ps @ z ky ky mm |} TOOUOOTIT = THOTT @ o Fig. 14.16 Springs connected in parallel. Figs. 14.16(a) and (b) show two springs connected in parallel. Let k and k, be their spring constants. Let y be the extension produced in each spring. Restoring forces produced in the two springs will be F,=—k,y and F, The total restoring force is FoF + Fy=-(ky +hy)y (1) Let k,, be the force constant of the parallel combi- nation. Then ky Fe-ky (2) From (1) and (2), k, =k +k Frequency of vibration of the parallel combination (Eee Voom Springs connected in series. Figs. 14.16(c) and (4) represent two springs connected in series. Let x, and x be the extensions produced in the two springs. The restoring force F acting in the two springs is same. is k, 2a Vm --(3) Letk, be the force constant of the series combination. Then F=-kx (4)14.22 PHYSICS-XI Kk Rte Frequency of oscillation of the series combination is From (3) and (4), FORMULAE USED 1. Spring factor or force constant, k 2. Period of oscillation of a mass m suspended from massless spring of force constant k, im Te2n |" k 3. For two springs of springs factors & and k, connected in parallel, effective spring factor, ke T=2n kak +k 4. For two springs connected in series, effective spring factor k is given by 5. When length of a spring is made n times, its spring factor becomes 1/n times and hence time period increases vin times. 6. When a spring is cut into n equal pieces, spring factor of each part becomes mk. 7m T22n [™ - nk Units Useo Spring factors k, k, k, arein Nm, mass mrisin kg and time period T in second, EXAMPLE 30. The pan attached to a spring balance has a mass of Tkg. A weight of 2 kg when placed on the pan stretches the spring by 10 cm What is the frequency with which the empty pan will oscillate ? Solution. Applied force, x98N=19.6N =2 kgwt= Displacement, y=10cm=0.1m pe F198 y OL For the empty pan, m=1kg Force constant, 196 Nm! Hence the frequency of oscillation of the empty pan will be go (El [oe 2nVm 2xV1 «21 4 me2 He Qn x Exaurst 31. A spring compressed by 0.2 m develops a restoring force of 25 N. A body of mass 5 kg is placed over it. Deduce : (i) force constant of the spring (ii) the depression of the spring under the weight of the body and . (iit) the period ofoscliation, ifthe body is disturbed Take g =10 N kg”. Solution. (i) Here y=0.2 m, F=25N Force constant, Fe y Om (i Here F =5 kg wt =5% 10 N=50N + Displacement, fale Es (iii) Here m=5 kg, k =125 Nm™ Time period, Ta2n | Qn Jo = 28s, k 15 «65 A0.2 kg of mass hangs at the end of a spring. Wier 0.02 kg more mass is added to the end of the spring, it stretches 7 cm more. If the 0.02 kg mass is removed, what will be the period of vibration of the system ? [Central Schools 04] Solution. When 0.02 kg mass is added, the spring stretches by 7 cm As mg=kx 1g _ 0.02 x 10 ee rei a 125 Nm 0.4m. When 0.02 kg. mass vibration will be int te Taam | =2n [02 "Ve? 2077 Exams: 33. A body of mass 12 kg is suspended by a coil spring of natural length 50 cm and force constant 2.0 10° Nov”, What is the stretched length of the spring ? removed, the period of 66.OSCILLATIONS 14.23 If the ext is pulled dono further stretching the spring toa length «of 59-cm and then released, what is the frequeney of escill- ation ofthe suspended mass ? (Neglect the mass of the spring). Solution. Here m=12 kg, k =2.0« 10° Nm~! Natural length, / = 50 em Extension produced in the spring due to 12 kg mass, Ee ee aaa =0.0588 m Stretched length of the spring = 1+ y =50 + 5.88 = 55.88 cm. When the loaded spring is further stretched, its fre- quency of oscillation does not change and is given by Tie E x10° m 2 a = 2.06 Hz. anVm 2x34 EAM sig, An impulsive force gives an initial velocity of =1.0 ms"? to the mass in the unstretched spring position [sce Fig. 14.17(a)]. What is the amplitude of motion ? Give x as a function of time t for the oscillating mass. Given m=3 kg and k =1200 Nav * Fig. 14.17 (a) Solution. Here initial velocity in unstretched position, v=-10mst Clearly, v,,, =1.0 ms"! Also, o= fe 21200 20 rads? m V3 Amplitude, A 101 =5em. a 20 20 As the motion starts from the unstretched position, the expression for the displacement can be written as sin wt =5 sin 20¢ As initial impulse is negative, the displacement is towards negative X-axis. So x=-5sin 20t. A 5kg collar is attached to a spring of force constant 500 Nur '. It slides without friction on a hori- zontal rod as shown in Fig. 14.17(b). The collar is displaced ‘from its equilibrium position by 10.0 cm and released. x= Examen 25kg Collar Fig. 14.17 (b) Calculate (i) the period of oscillation, (if) the maximum speed, and (iii) the maximum acceleration of the collar. : INCERT ; Dethi 03C] 500 Nm=!, Solution. Here m=5kg, k A=10.0 cm =0.10 m () Period of oscillation, T=2 x [it-28 = k 500 =23.14% 1 5 -0.628s. 10 (ii) The maximum speed of the collar, », -oa- [ae 500 0.10 a mons 1 =10ms7. (ii) The maximum acceleration of the collar, dings = 0? A=* A= 550.10 =10 ms. " nda) ExAupit: 36. A small trolley of mass 2.0 kg resting on a horizontal turn table is connected by a light spring to the centre of the table. When the turn table is set into rotation at a speed of 300 rpm. the length of the stretched spring is 40 con If the original length of the spring is 35 cm determine the force constant of the spring Solution. Mass of trolley, m=2.0 kg Frequency of rotation of turn table, oa aT Extension produced in the string, y=40-35=5 cm=5x 10°? m When the turn-table is set into rotation, the tension (restoring force) in spring is equal to the centripetal force. Thus Restoring force = Centripetal force F =ky=mrot =mr(2nvy 4x? v? mr y [r=length of stretched spring = 40 cm] 4% 9.87 x 5? «2.0 40 x 10°? . 5x10? or k = 15795 Nm NAMELY 37. Two masses m, =1.0 kg and mb =05 kg are Suspended together by a massless spring of force constant, k=12.5 Nur |, When they are in equilibrium position, m, is gently removed. Calculate the angular frequency and the ‘amplitude of oscillation of m,. Given ¢ =10 ms~14.24 PHYSICS-XI Solution. Let ybe the extension in the length of the spring when both m and nm are suspended. Then F=(m,+m)¢=ky (om +m) 8 k Let the extension be reduced to y when 1m, is removed, then or y 7 yas Fig. 14.18 (m+ mds ms k k k This will be the amplitude of oscillation of m, x10 Amplitude, A=" 0.8 m. Angular frequency, oe [5 JS og te. ym Vos Two identical springs, each of spring factor k, ‘may be connected in the following ways. Deduce the spring {factor of the oscillation of the body in each case. Solution. For each spring, ky (1) where _F = restoring force, k = spring factor, and y= displacement of the spring. LJ» k k Fig. 14.19 (i) In Fig, 14.19(@), let the mass m produce a displacement y in each spring and F be the restoring force in each spring. If k, be the spring factor of the combined system, then 2F=-ky or (2) (i) In Fig, 14.19(0), as the length of the spring is doubled, the mass mwill produce double the displace- ment (2y). If k, be the spring factor of the combined system, then F = ~kyQy)=-2ky 8) ‘Comparing (1) and (3), k 2ky=k oF ky =a, 2 2 (iii) In Fig. 14.19(c), the mass m stretches the upper spring and compresses the lower spring, each gi rise to a restoring force F in the same direction. If k, be the spring factor of the combined system, then (4) Two identical springs, each of force constant K are connected in (a) series (b) parallel, and they support a ‘mass m Calculate the ratio ofthe time periods of the mass in the two systems. [Central Schools 07] Solution. (2) For series combination, the effective force constant is kek _k eek 2 T,=2n|™ =2n | ™ K, K (#) For parallel combination, the effective force constant is k,OSCILLATIONS _ 14.25 rv A tray of mass 12 kg is supported by tewo = 5 ‘al springs as shown in Fig. 14.20. When the tray is Lroraoor! 7 -osoon| pressed down slightly and i i released, itexecutes SHM with tx 4 time period of 1.5 s. What is the force constant of each J. 14.22 spring? When a black of mass Force exerted by the right spring, M is placed on the tray, the Fy =~, towards lef period of SHM changes to os i 30's What is the mass of the ‘The net force acting on mass mis block ? Fig. 14.20 FoF, + hy=-2ke Solution. Let k be the force constant of each spring. As the two springs are connected in parallel, so the force constant of the combination is =k+k=2k [m wo dT =2n No me m_4x (3.14) x a k 4x? im _ 4x B14)? x 12 7 10.34 Nm k= k/2 = 105.17 Nm. When a block of mass M is placed in the tray, the period of oscillation becomes sy M+ m T « Ke Mim M+i2 ™ 12 M+12 M= 48-12 =36 kg. wis: 41. The identical springs of spring constant k are lttached to a block of mass m and to fixed supports as shown below. |-avotn0e [wove Fig. 14.21 Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find the period of oscillations, INCERT] Solution. As shown in Fig. 14.22, suppose the mass mis displaced by a small distance x to the right side of the equilibrium position O. Then the left spring gets elongated by length x and the right spring gets compressed by the same length x. Force exerted by the left spring, F, =—kx, towards left ‘Thus the force acting on the mass mis proportional to its displacement x and is directed towards its mean position. Hence the motion of the mass mis simple harmonic. Force constant is =2k ‘The period of oscillation is m 7 anf k m 2k Exansy 42. A trolley of mass3.0 kg is connected to too identical springs each of force constant 600 Nui” ', as shown in Fig, 14.23. If the trolley is displaced from its equilibrium position by 5.0 cm and released, what is (i) the period of ensuing oscillations, (ii) the maximum speed of the trolley ? (iii) How much is the total energy dissipated as heat by the time the trolley comes to rest due to damping forces ? T=2n ooNm' SOK coo nm fF DOTHOTT Fig. 14.23 Solution. When the trolley is displaced from the mean position, it stretches one spring and compresses the other by the same amount. The restoring forces developed in the two springs are in the same direction. If the trolley is displaced through distance y, then total restoring force is, Poa eons Ifk’ is the force constant of the combination, then Fe-ky Clearly, k’ =2k =2 x 600 =1200 Nm=" Also, 0 kg amplitude, A=5.0x 10"? m (i Period of oscillation, 0314s.14.26 PHYSICS-xI (ii) Maximum speed, x 5.0 107? =1.0 ms“ (ii) Total energy dissipated as heat = Initial maximum K.E. of the trolley =) im? =} 90% (107 =15}. 2 Mma X% PROBLEMS For Practice 1. A spring compressed by 0.1 m develops a restoring force of 10 N. A body of mass 4 kg is placed on it. Deduce (i) the force constant of the spring (i) the depression of the spring under the weight of the body and (iii) the period of oscillation, if the body is disturbed. Ans. () 100 Nev“ (ji) 0.4 m (ili) 1.26 5} 2. The period of oscillation of a mass m suspended by an ideal spring is 2s. If an additional mass of 2 kg be suspended, the time period is increased by 1 s. Find the value of m (Ans. 16 kg) 3. An uncalibrated spring balance is found to have a period of oscillation of 0.314 s, when a 1 kg weight is suspended from it ? How does the spring elongate, when a 1 kg weight is suspended from it ? [Take x = 314] (Ans. 2.45 cm) The frequency of oscillations of a mass m suspended by a spring is v,. If the length of the spring is cut to one-half, the same mass oscillates with frequency v.. Determine the value of v» /v,. {Chandigarh 03] (Ans. 2) 5. The periodic time of a mass suspended by a spring (force constant f) is T. If the spring is cut in three ‘equal pieces, what will be the force constant of each part ? If the same mass be suspended from one piece, what will be the periodic time ? (Ans. 3k, T/ V3) 6, The time period of a body suspended by a spring be T. What will be the new period, if the spring is cut into two equal parts and when (j) the body is suspended from one part (i) the body is suspended from both the parts connected in parallel. [Ans. (i) T/ V2 (ii) T/2] 7. Two identical springs have the same force constant of 147 Nm~", What elongation will be produced in each spring in each case shown in Fig. 14.24? Take g = 9.8 ms~. (Ans. (a) 1/6 m (@) 1/3 m, 1/3 m (¢) 1/3 m] 2 Tig Ske K k sks © © © Fig. 14.24 Fig. 14.25 8. Three springs are connected to a mass mas shown, in Fig. 14.25. When mass mm oscillates, what is the effective spring constant and time period of vibration ? Given k= 2Nm~! and m= 80g, (Ans. 8 Nm", 0,628 s) 9. Two springs are joined and connected to a mass m as shown in Fig, 14.26. If the force constants of the two springs are k, and ky, show that frequency of oscillation of mass mis 2 [ie “On YR +e) mt ky ci Fig. 14.26 % HINTS 10 F 1 a) k= F=1L00Nm OF or x10 100 2.2, fa winrar fF nae fig tate : In first case, (i) yah 420 04m, (1)Dividing (2) by (1), we get 9_m+2 8 1. om oe 5 1. Let k be the force constant of the full spring. Then frequency of oscillation of mass m will be a fe an Vn When the spring is cut to one-half of its length, its force constant is doubled (2k). Frequency of oscilla tion of mass m will be % 2 vgly =v 16 kg. 2n . Time period of mass m when suspended from the full spring is peo ft k When the springis cut into three equal parts, the force constant of each part becomes 3k. Time period of _mass mwhen suspended from one such piece will be m_T Tare (7 "3K 3 a IF the spring is cut into two equal parts, then the force constant of each part becomes 2k. . For full spring, T= 2x (@) When the body is suspended from one part, its period of oscillation is mT. refi (ii) For the two parts connected in parallel, force constant = 2k + k= tk ‘The period of oscillation becomes i r . Here k= 147 Nm-~'. In Fig. 14.24(a), the effective spring constant, Kak+k=2k=2147=294Nm + Elongation in the spring, mg _ 5x93 K "234 In Fig. 14.24(b), the effective spring constant, kxk _k_ 147 "Thar 27 2 ~ Total elongation in the spring, w5x92%2 2 a o Nm“! Ya Elongation nach pring =m OSCILLATIONS 14.27 In Fig. 14.24(0), the effective spring constant, K=147Nm* : 5x98 1 Elongation in the spring, y, = -~2% gation in the spring. Y= Te = m. The given arrangement is equivalent to the three springs connected in parallel. The effective spring constant is Kak+ 2+ k= Time period, Toa [Haan 008 OVE 7 V8 9. Let a force F applied on the body produce displacements x, and x, in the two springs. Then 4x2=8Nm7, 628 s. Clearly, force constant of the system, k= 4 Rte Viv [ati 14.16 — SIMPLE PENDULUM 19. Show that for small oscillations the motion of a simple pendulum is simple harmonic. Derive an expression or its time period. Does it depend on the mass of the bob ? Simple pendulum. An idea! simple pendulum consists ofa point-mass suspended by a flexible, inelastic and weightless string from a rigid support of infinite mass. In practice, we can neither have a point-mass nor a weightless string. In practice, a simple pendulum is obtained by suspending a small metal bob by a long and fine cotton thread from a rigid support. Expression for time period. In the equilibrium position, the bob of a simple pendulum lies vertically below the point of suspension. If the bob is slightly displaced on either side and released, it begins to oscillate about the mean position. Suppose at any instant during oscillation, the bob lies at position A when its displacement is OA =x and the thread makes angle @ with the vertical. The forces acting on the bob are (0. Weight mg of the bob acting vertically downwards. (ii) Tension T along the string. Frequency, v=14.28 PHYSICS-XI of suspension mg cos mg Fig. 14.27 Force acting on the bob of a pendulum. ‘The force mg has two rectangular components (i) the component mg cos @ acting along the thread is balanced by the tension T in the thread and (ii) the tangential component mg sin 0 is the net force acting on the bob and tends to bring it back to the mean position. Thus, the restoring force is eee F =-mg sin @=-mg| 9-24 2... i s(' 315i ote =-mge{1-2 42 so 6 "120 where @ is in radians. Clearly, oscillations are not simple harmonic because the restoring force F is not proportional to the angular displacement 0. However, if @ is so small that its higher powers can be neglected, then F=-mg6 If lis the length of the simple pendulum, then ota (a eadus 7 mg * 87 or or. ‘Thus, the acceleration of the bob is proportional to its displacement x and is directed opposite to it. Hence for small oscillations, the motion of the bob is simple harmonic. Its time period is or T=2n. iE 8 2x g/t Obviously, the time period of a simple pendulum depends on its length land acceleration due to gravity ¢ but is independent of the mass m of the bob. @ Examples based on Formutar Use Tr 1. Time period, T= 26 7 8 uf Length | of the pendulum is in metre and acceleration due to gravity g in ms, 2. Frequency, v = Units Useo Ewaurin a. What is the length of a simple pendulum, which ticks seconds ? INCERT ; Delhi 09] Solution, The simple pendulum which ticks seconds a second pendulum whose time period is 2 s. Thus T=25, g=98ms* AMPLi44. A pendulum clock shows accurate time. If the length increases by 0.1%, deduce the error in time per day. {Delhi 951 Solution. Correct number of seconds per day, Vv =24 x 60 x 60 =86400. Let error introduced per day =x seconds Then incorrect number of seconds per day, v =86400 + x If lis the original length of the pendulum, then its new length will be P=1+0:% of 1=1+ 22%! =(1+ 0.001)! 100 Now frequency, v if jen var tf (eee (caer v Vr 86400 \(1+0.001)7 x or (1+0.001) "? $0001 =1 ~0.0005 or 0.0005 0.0005 x 86400 = - 43.2. The negative sign shows that the clock will run slow and it will lose 43.2 seconds per day. orOSCILLATIONS 14.29 Examrrr 45. Two pendulums of lengths 100m and (iii) When the carriage moves horizontally, both g 110.25 omstart oscillating in phase. After how many oscilla- and a are at right angle to each other, hence the net tions will they again be in same phase ? acceleration is Solution. The two pendulums will be in same a= \ gee? = oay+ (ay phase again when large pendulum completes v oscilla- tions and small pendulum completes (v + 1) oscillations. = (96.04 + 16 = /112.04 =10.58 ms~? For larger pendulum, Time period will be a fi ae F on (re 2nVI Qn VII025 7x 1058 1 [a For smaller pendults vere d [ie =2% 096 =1.928, F smaller pendulum, 35 Vi00 2x 0.96 =1.92s. ‘xanpet 47. The bottom of a dip on a road has a radius of Ccuroature R. A rickshaw of mass M lefta little away from the bottom oscillates about the dip. Deduce art expression for the 25 ) period of oscillation. [Chandigarh 02) 00, Solution. As shown in Fig. 14.28, let the rickshaw. ‘of mass Mbeat position A at any instant and ZAOB= 6. or or ‘Thus the two pendulums will be in same phase when the larger pendulum completes 20 oscillations or smaller pendulum completes 21 oscillations. Examvt: 46. A second’s pendulum is taken in a carriage. Find the period of oscillation when the carriage moves with an acceleration of 4 ms”? (i) vertically upwards (ii) verti- cally downwards, and (iii) in a horizontal direction. Solution. Time period of a pendulum, Tran ft el For second’s pendulum, T aan or 8 Fig. 14.28 Forces acting on the rickshaw at position A are or eu (i) Weight Mg acting vertically downwards. 8 (i) The normal reaction N of the road. (®) When the carriage moves up with an accele- The weight Mg can be resolved into two ration a=4 ms~?, the time period is rectangular components : n mee 9. (i. Mg cos 6 perpendicular to the road. It balances Sera” "| Foae® the normal reaction N. 2n [98 (i) Mg sin @ tangential to the road. It is the only Sls eee unbalanced force acting on the rickshaw which : acts towards the mean position B Hence the (i) When the carriage moves down with an restoring force is acceleration a =4 ms~?, the time period is Fe—Mgsin® 7 os Arc _ AB T,=2n | z— in = 9=—Ate_ -AB_¥ 2 7 i 08-4) For small, sin@=0=, 0 == a8 Mg y = =2«1.299= "By ie, Foe 22 [25 =2 «1299-2595, 7 gt14.30 PHYSICS-x! Hence the motion of the rickshaw is simple harmonic with force constant, an Mf Time period, (mM, [a T=2n,|M- =2n 2 Ve aaa RY X% PROBLEMS FoR PRACTICE 1. The time taken by a simple pendulum to perform 100 vibrations is § minutes 9 seconds in Bombay and 8 minutes 20 seconds in Pune. Calculate the ratio of acceleration due to gravity in Bombay and Pune. (Ans, 1.0455) 2. Ifthe length of a pendulum is decreased by 2%, find the gain or loss in time per day. (Ans. Gain of 864 5) 3. Ifthe length of a second’s pendulum is increased by 1%, how many seconds will it lose or gain ina day ? (Ans. Loss of 432 s) 4. If the length of a simple pendulum is increased by 45%, what is the percentage increase in its time period ? (Ans. 22.5%) 5. What will be the time period of second’s pendulum if its length is doubled ? (Ans. 2.828 5) 6. Ifthe acceleration due to gravity on moon is one-sixth of that on the earth, what will be the length of a second pendulum there ? Take g = 9.8 ms" *. (Ans. 16.5 cm) Xx HINTS 1. Let g, and g, be the values of acceleration due to gravity in Bombay and Pune and T; and T, be the values of the time-periods at the respective places. Then 8min9s_ 489 i = 88 nas #00 * 100° = 4898 Smin 208 _ 5008 _ 5 100 100” As 8 eo EF 0485. 7 (4.897 2. Asv.« 1/41, so the number of seconds gained per day on decreasing the length by 2% Al 12 * <86400= <2 «86400 = 864 s. ; op * 86400 = B64 s 1 2 F— con increasing the length by 45% =141 100-1 21 6. On the moon, g,,= 45 Let the tube be depressed in water by a little distance y and then released. Spring factor, F_Ay.p. Examen 50. A cylindrical wooden block of cross-section 15.0 ent? and mass 230 g is floated over toater with an extra weight of 50 g attached to its bottom. The cylinder floats vertically. From the state of equilibrium, it is slightly depressed and released. If the specific gravity of wood is 0.30 and 18 ms~?, deduce the frequency of the block, A=15.0 on? =15 10" Solution. Area of cross-section of the block, A=15.0 cm? =15x 10-* m? Total mass of the block, m=230 + 50 =280 g Density of water, o=10° kg m= Density of wood, p =0.30 x 10° kg m™?=300 kg m™? 1.28 kg, 3 OSCILLATIONS 14.33 Ani5em Cylinder Fig. 14.34 Let the cylinder be depressed through a small distance y. Then Restoring force = Weight of water displaced F=Ayos Force constant, or k === Aog=15x10"4 «10°98 y 1 fk _ 1 fia7 Fi ive |i 4 POUENEY 25 Yi ™3 5 VO z =“ x J525 = 1.15 Hz. 4 Exams 5t. The balance wheel of a watch has a moment of inertia of 2 x 10~* kg n?* and the torsional constant of its hair spring is 9.8x10"° Nmrad~'. Calculate its frequency. Solution, Here 1 =2« 10~* kgm”, C=9.8% 10° Nm rad! Frequency, if 70 _7%317 2nVI 2x314 2x314 Exams 52. A sphere is hung with a wire. 30° rotation of the sphere about the wire generates a restoring torque of 4.6 Nm If the moment of inertia of the sphere is 0.082 kg nt, deduce the frequency of angular oscillations. =3.53 Hz. Solution. Here 0 =30°=7 rad, t= 4.6 Nm, 1 =0.082 kg m? Restoring torque per unit angular displacement, 6 =8.78 Nm rad!14.34 PHYSICS-XI X PROBLEMS FOR PRACTICE 1. Ifthe earth were a homogeneous sphere and a straight hole was bored in it through its eentre, show that a body dropped in the hole SHM and cal- culate the time period ofits vibration. Radius of earth is 64% 10° mand ¢=9.8ms"*, (Ans. 5077.6 s) 2. A weighted glass tube is floating in a liquid with 20 em of its length immersed. It is pushed down through a certain distance and then released. Show that up and down motion executed by the glass tube is SHM and find the time period of vibration. Given, ¢ = 980em s°?, (Ans. 0.898 s) 3. A sphere is hung with a wire. 60° rotation of the sphere about the wire produces a restoring torque of 4.1 Nm. If the moment of inertia of the sphere is 0.082 kg m?, find the frequency of angular oscillations. (Ans. 1.1 Hz) 4. A lactometer whose mass is 0.2 kg is floating vertically in a liquid of relative density 0.9. Area of cross-section of the marked portion of lactometer is, 0.5% 10 m2, If it is dipped down in the liquid slightly and released, what type of motion will it execute ? What will be its time-period ? (Ans. Motion is simple harmonic, 4.2 s) x HINTS 1, Here R=64%10° m, ¢=98m oulitin rome ff (ae 8 2 Here h=20cm, g = 980cms* i 20 =2n |" =2n | 2 = 25 «0.143 = 0.898 s. Ta2ndf = 2 aap 0.143 = 0.8 = 2n x 808.1 = 5077.6 s. 3. Restoring torque, t= 4.1 Nm Angular displacement, 0 = 60° 2) 41 413 mn constant, 41 2813 Nim rad Torsion constant, C= 5 = 7 = Moment of inertia, | = 0.082 kg m? 1 fe _1 [3x41 Frequency, v= [© Fe Vecooeg =ht He 4, When the lactometer is depressed through distance y, F = upthrust of the liquid =~ Ayp xg=-Apgy As F «y, so motion of lactometer is SHM with =425, x09x10° x98 14.18 FREE, DAMPED AND MAINTAINED OSCILLATIONS 25. What are free, damped and maintained oscillations ? Give examples. (a) Free oscillations. Ifa body, capable of oscillation, is slightly displaced from its position of equilibrium and left to itself, it starts escillating with a frequency of its own. Such oscillations are called free oscillations. The frequency with which a body oscillates freely is called natural frequency and is given by wee 0° an Vm Some important features of free oscillations are (i) Inthe absence of dissipative forces, such a body vibrates with a constant amplitude and fixed frequency, as shown in Fig. 14.35. Such oscil- lations are also called undamped oscillations. (ti) The amplitude of oscillation depends on the energy supplied initially to the oscillator. (iii) The natural frequency of an oscillator depends on its mass, dimensions and restoring force ie., on its inertial and elastic properties (mand k). Fig. 14.35 Free or undamped oscillations. Examples. (i) The vibrations of the prongs of tunning fork struck against a rubber pad. (i?) The vibrations of the string of a sitar when pulled aside and released. (iii) The oscillations of the bob of a pendulum when displaced from its mean position and released, (b) Damped oscillations. The oscillations in which the amplitude decreases gradually with the passage of time are called damped oscillations. In actual practice, most of the oscillations occur in viscous media, such as air, water, etc. A part of the energy of the oscillating system is lost in the form of heat, in overcoming these resistive forces. As a result, the amplitude of such oscillations decreases exponen- tially with time, as shown in Fig. 14.36. Eventually, these oscillations die out.In an oscillatory motion, friction produces three effects (i) It changes the simple harmonic motion into periodic motion. (ii) It decreases the amplitude of oscillation. (iii) It slightly reduces the frequency of oscillation Fig. 14.36 Damped oscillations. Examples. (i) As shown in Fig. 14.37, consider a block of mass nt that oscillates vertically on a spring, with spring constant k. The block is connected to a vane through a rod. The vane is submerged in a liquid. lates up and down, the vane also oscillates in a similar manner inside the liquid. The liquid exerts an opposing force of viscosity on the vane. The energy of the oscillating system is lost in the liquid as heat. The amplitude of oscillation decreases continuously with time. Rigid support Scale Springiness, k Mass, —Vane ‘damping, b Fig. 14.37 A damped simple harmonic oscillator. (ii) The oscillations of a swing in air. (iii) The oscillations of the bob of a pendulum in a fluid. (©) Maintained oscillations. If to an oscillating system, energy is continuously supplied from outside at the same rate at which the energy is lost by it, then its amplitude OSCILLATIONS _ 14.35 can be maintained constant. Such oscillations are called ‘maintained oscillations. Here, the system oscillates with its own natural frequency. Examples. (i) The oscillations of the balance wheel of a watch in which the main spring provides the required energy. (ii) An electrically maintained tuning fork. (iii) A child’s swing in which energy is continuously fed to maintain constant amplitude. 4 Differential equation for damped oscillators and its Solution. In a real oscillator, the damping force is proportional to the velocity v of the oscillator. F,=— bv where b is damping constant which depends on the characteristics of the fluid and the body that oscillates in it, The negative sign indicates that the damping force opposes the motion. Total restoring force =~ kx - bu ax dx dx or mE aie -b a "ae dt | 2 or TF ob + keno This is the differential equation for damped S.H.M The solution of the equation is x()=a.e7§" cos(a't + 4) The amplitude of the damped S.H.M. is a’= ae" Y/2m ‘where a is amplitude of undamped S.H.M. Clearly, a” decreases exponentially with time, ‘The angular frequency of the damped oscillator is Clearly, damping increases the time period (due to the presence of the term b? /4 m? in the denominator). ‘The mechanical energy of the damped oscillator at any instant 1 will be Ew} 1 J gg? = gg? ebm 5 ae Obviously, the total energy decreases exponentially time. ‘As damping constant, b=F/o N_ _kg ms ee Sl unit of b= kgs ms CGS unit of b=g 57.14.36 PHYSICS-x! EXAMPLE 53. For the damped oscillator shown in earlier Fig, 14.37, the mass m of the block is 200 ¢ 90 Nin! and the damping constant b is 40 gs". Calculate (a) the period of oscillation, (b) tinte taken for its amplitude of vibrations to drop to half of its initial vatue and (c) the time taken for its mechanical energy to drop to half its initial value. INCERT] Sol (90 x 0.200 = 4.24 kgs! n. (a) Here Vk Damping constant, b= 40 gs" As the damping constant, b << Jkm, is small, so the time period T is given by T=2n[ 22n Oz k 90 Nm~ 03s. (b) The time, 7, . for the amplitude to drop to half of its initial value is given by AL jg @hyavem 2 (©) The time, t,/» for the mechanical energy to drop to half its initial value is given by E (typ) = Eye 2 or E (tyy_)/ E(O)= exp (~ by» /m) or 1/2 = exp (- btyj»/m) In (1/2) =~(btyyp /m) 0.693, or fa ip get *200 8-345 14.19 FORCED AND RESONANT OSCILLATIONS 26. Distinguish between forced and resonant oscillations. Give an experimental illustration in support of your answer. Give examples. Forced oscillations. When a body oscillates under the influence of an external periodic force, not with its own natural frequency but with the frequency of the external periodic force, its oscillations are said to be forced oscil- lations. The external agent which exerts the periodic force is called the driver and the oscillating system under consideration is called the driven body. Examples. (i) When the stem of a vibrating tuning fork is pressed against a table, a loud sound is heard. This is because the particles of table are forced to vibrate with the frequency of the tuning fork. (ii) When the free end of the string of a simple pendulum is held in hand and the pendulum is made to late by giving jerks by the hand, the pendulum. executes forced oscillations. Its frequency is same as that of the periodic force exerted by the hand, (iii) The sound boards of all stringed musical nstruments like sitar, violin, etc. execute forced scillations and the frequency of oscillation is equal to the natural frequency of the vibrating string, Suppose an external periodic force of frequency vis applied to an oscillator of natural frequency, Initially, the body tries to vibrate with its own natural frequency, while the applied force tries to drive the body with its own frequency. But soon the free rations of the body die out and finally the body vibrate a with a constant amplitude and with the frequency of the driving force: In this steady state, the rate of loss of energy through friction equals the rate at which energy is fed to the oscillator by the driver, Fig, 14.38 shows the variation of the amplitude of forced oscillations as the frequency of the driver varies from zero to a large value. Clearly, the amplitude of forced oscillations is very small for v <> vy But when v=vy the amplitude of’ the forced oscillations becomes very large. In this condition, the oscillator responds most favourably to the driving force and draws maximum energy from it. The case v =v, is called resonance and the oscillations are called resonant oscillations, Fig. 14.38 Amplitude a of a forced oscillator as a function of the frequency v of the driver. Resonant oscillations and resonance. It is a particular case of forced oscillations in which the frequency of the driving force is equal to the natural frequency of the oscillator itself and the amplitude of oscillations is very large. Such oscillations are called resonant oscillations and phenomenon is called resonance, Examples. (i) An aircraft passing near a building shatters its window panes, if the natural frequency of the window matches the frequency of the sound waves sent by the aircraft's engine. (ii) The air-column in a reasonance tube produces a loud sound when its frequency matches the frequency of the tuning fork. (iii) A glass tumbler ot a piece of china-ware on shelf is set into resonant vibrations when some note is sung or played.Experimental illustration. As shown in Fig. 14.39, suspend four pendulums A, B, Cand D from an elastic string PQ. Set the pendulum A into oscillation. It executes free oscillations. The energy from this pendulum is transferred to other pendulums through the elastic string, Initially, the motions of B Cand Dare irregular, But’ soon all these pendulums start oscillating with the frequency of A. The oscillations of B,C and D are forced oscillations. But pendulums B and D have small amplitudes. This is because the frequency of Bis much larger than that of A (due to shorter length) and the frequency of Dis much smaller than that of A (due to larger length). The pendulum C which has same length as the pendulum A (and hence the same frequency) oscillates with largest amplitude. ations of C are resonant oscillations. Elastic strin P. 6 Q d 14,39 Illustrating free, forced and resonant oscillations. 27. Briefly explain the principle underlying the tuning of a radio receiver. Principle of tuning of a radio receiver. Tuning of the radio receiver is based on the principle of resonance. Waves from all stations are present around the antenna. When we tune our radio to a particular station, we produce a frequency of the radio circuit which matches with the frequency of that station. When this condition of resonance is achieved, the radio receives and responds selectively to the incoming waves from that station and thus gets tuned to that station. OSCILLATIONS _ 14.37 14.20 COUPLED OSCILLATIONS 28. What are coupled oscillations ? Give examples Coupled oscillations. A system of ttoo or more oscil~ ators linked together in such a way that there és nnutual exchange of energy between thent is called a coupled oscillator. The oscillations of such a system are called coupled oscillations. Examples. (i) Two masses attached to each other by three springs between two rigid supports. The middle spring provides the coupling between the driver and the driven system [Fig. 14.40(a)]. (ii) Two simple pendulums coupled by a spring (Fig. 14.40(6)]. (iii) Two LC-circuits placed close to each other. The circuits are linked by each other through the magnetic lines of force (Fig. 14.40()} k @ ” © Fig. 14.40 Coupled oscillators. ‘When two identical oscillators are coupled together, the general motion of such a system is complex. It is periodic but not simple harmonic. It can be viewed as the superposition of two independent simple harmonic motions, called normal modes having angular frequencies ; and o,. The constituent oscillators execute fast oscillations of average angular frequency, @,, = (00, + 0,)/2. The amplitude of either oscillator varies with an angular frequency («,-0,) This phenomenon of variation of amplitudes is known as beats and the frequency («, ~ «,)is called beat frequency Very Short Answer Conceptual Problems Problem 1, Can a motion be periodic and not oscillatory ? Solution. Yes. For example, uniform circular motion is periodic but not oscillatory. Problem 2. Can a motion be oscillatory but not simple harmonic ? If your answer is yes, give an example and if not, explain why. Solution. Yes; when a ball is dropped from a height on. a perfectly elastic surface, the motion is oscillatory but not simple harmonic as restoring force F = mg =constant and not F oc~ x, which is an essential condition for S.H.M. Problem 3. Every simple harmonic motion is periodic motion, but every periodic motion need not be jimple harmonic motion. Do you agree ? Give one example to justify your statement.14,38 PHYSICS-xI Solution. Yes, every periodic motion need not be simple harmonic motion. For example, the motion of the earth round the sun is a periodic motion, but not simple the back and forth motion is: not harmonic motion a taking place. Problem 4. The rotation of the earth about its ax periodic but not simple harmonic. Justify. Solution. The carth takes 24 hours to complete its rotation about its axis, but the concept of to and fro motion is absent, and hence the rotation of the earth is, periodic and not simple harmonic. Problem 5. What is the basic condition for the motion of a particle to be S.H.M.? [Delhi 02) Solution. The motion of a particle will necessarily be simple harmonic if the restoring force acting on it is proportional to its displacement from the mean position ie, P=-kx. Problem 6. Which of the following conditions is not sufficient for simple harmonic motion and why ? (i) acceleration « displacement, (ii) restoring force « displacement. Solution. Condition (i) is not sufficient because it does not mention the direction of acceleration. In S.H.M. the acceleration is always in a direction opposite to that of the displacement. Problem 7. Are the functions tan ot and cot ot periodic ? Are they harmonic ? Solution. Both tant and cot wt are periodic is functions each with period T= x/« because ( tan{ o(1 + 2)] = tan (ot +») =tan ot ow, cat] o(s +3) -cot tot + 2)=cot ot p But these functions are not harmonic because they can take any value between 0 and «. Problem 8. What provides the restoring force for simple harmonic oscillations in the following cases : (@ Simple pendulum (ii) Spring (iii) Column of Hg in U-tube ? Solution. (i) Gravity (ii) Elasticity (iii) Weight of difference column Problem 9. When are the displacement and velocity in the same direction in S.H.M. ? and Solution. When a particle moves from mean position to extreme position, its displacement and velocity are in the same direction. Problem 10. When are the velocity and acceleration, in the same direction in S,H.M. ? Solution. When a particle moves from extreme position to mean position, its velocity and acceleration are in the same direction, Problem 11. Can displacement and acceleration be in the same direction in S.H.M. ? Solution. No. In S.H.M., acceleration is always in the opposite direction of displacement. Problem 12, The relation between the acceleration a and displacement x of a particle executing S.H.M. is = ~(plq) y, where p and q are constants. What will be the time period T of the particle ? P 4 Solution. Here @ = oy, where @= ©. Vp Problem 13. The maximum acceleration of a simple harmonic oscillator is a, and the maximum velocity is, Ug. What is the displacement amplitude? ¢in; 99) Solution, Let A be the displacement amplitude and be the angular frequency of S.H.M. Then Maximum velocity, m= 0A o 2 Maximum acceleration, ay = 0? A = (2) A 0 Problem 14. The time period of an oscillating body is given by T=2pJmladg . What would be the force equation for the body ? Solution. On comparing the given equation T =2n m/adg with the standard equation T = 2x Jm/k, we get k = adg, which gives the force equation F = — adg (y). Problem 15. Two simple pendulums of unequal length meet each other at mean position while oscillating. What is their phase difference ? Solution. If both pendulums are moving in the same direction, then $= 0° and if they are moving in opposite directions, then 6 = 180° or x radian. Problem 16. Velocity and displacement of a body executing $.H.M. are out of phase by x/2. How ? Solution, Displacement, x =a.cos wt dr Velocity, » asin ot = @acos (wt + x/2) dt Clearly, velocity leads the displacement by x/2 rad. Problem 17. A particle executes S.H.M. of amplitude A. At what positions of its displacement (x), will its (# velocity be zero and maximum and (ii) acceleration be zero and maximum ? Solution. (i) Zero velocity at x velocity at x = 0." (ii) Zero acceleration at x = 0, maximum acceleration at, A + A, maximumProblem 18. At what points along the path of a simple pendulum is the tension in the string (i) maximum and (if) minimum ? Solution. (i) The tension is maximum at the mean position and is equal to mg, where 1m is the mass of the bob. (ii) The tension is minimum at either extreme position and is equal to mg cos 0, where 0 is the angle through Which the string gets displaced to reach the extreme position. Problem 19. Is the statement “the bob of a simple pendulum moves faster at the lowest position for larger amplitude” true ? Justify your answer. Solution. We know that velocity of a simple pendulum is maximum at the lowest position (mean position) and is given by v max = OA ie. for larger amplitude (A), the bob of simple pendulum would move faster. Problem 20. Can we use a pendulum watch in an artificial satellite? Solution. No. In an artificial satellite, a body is in a state of weightlessness, i.e. ¢ = 0 T=2n £ Inside the satellite, the pendulum does not oscillate. Hence a pendulum watch cannot be used in an artificial satellite Problem 21. A girl is swinging in the sitting position. How will the period of the swing change if she stands up? IAIEEE 02 ; Certtral Schools 09] Solution. The girl and the swing together constitute a pendulum of time period, Tr roan | As the girl stands up, her C.G. is raised. The distance between the point of suspension and the C.G. decreases i.e, length I decreases. Hence the time period T decreases, Problem 22. Will a pendulum clock lose or gain time when taken to the top ofa mountain? [Himachal 04] Solution. On the top of the mountain, the value of g is, less than that on the surface of the earth. The decrease in the value of g increases the time period of the pendulum (on the top of the mountain. So the pendulum clock loses time, Problem 23. What will be the period of oscillation, if the length of a second’s pendulum is halved ? i_e 2x2 Solution. [+= 7 or a 2%? y 2 in B . B=2 o he=vis OSCILLATIONS 14.39 Problem 24. The length of a second’s pendulum on the surface of earth is 1 m. What will be the length of a second’s pendulum on the surface of moon ? ' Solution. T= 2x |! Vs In both the cases, T is same so that leg On the moon, the value of acceleration due to gravity, is one-sixth of that on the surface of earth. So the length of second’s pendulum is 1m, Problem 25. The bob of a simple pendulum is made of wood. What will be the effect on the time period if the wooden bob is replaced by an identical bob of iron ? Solution. There will be no effect because the time period does not depend upon the nature of material of the bob, Problem 26. Ifa hollow pipe passes across the centre of gravity of the earth, then what changes would take place in the velocity and acceleration of a ball dropped in the pipe? Solution. The ball will execute S.H.M. to and fro about the centre of the earth. At the centre, the velocity of the ball will be maximum (acceleration zero) and at the earth’s, surface the velocity will be zero (acceleration maximum). Problem 27. The bob of a simple pendulum of length | is negatively charged. A positively-charged metal plate is placed just below the bob and the pendulum is made to oscillate. What will be the effect on the time- period of the pendulum ? Solution. The positively charged metal plate attracts the negatively charged bob. This increases the effective value of g. Hence the time period will decrease. Problem 28. A simple pendulum of length ! and with a bob of mass m is moving along a circular arc of angle 0 ina vertical plane. A sphere of mass m is placed at the end of the circle. What momentum will be given to the sphere by the moving bob ? Solution. Zero. This is because the velocity of the bob at the end of the arc will be zero. Problem 29. A body moves along a straight lineOAB simple harmonically. It has at zero velocity at the points ‘A and B which are at distances a and b respectively from O and has velocity » when half way between them. Find the period of S.H.M. Solution. Clearly, Cis the mean position of S.H.M., as shown in Fig. 14.41 -/_———— » ——4 -— *# —4 — _ ° A Fig. 14.41,14.40 PHYSICS-XI The amplitude of S.HM Az AC ‘The velocity at the mean position C will be wade tt fae pe nb=a) Problem 30. When a 2.0 kg body is suspended by a spring, the spring is stretched. If the body is pulled down slightly and released, it oscillates up and down. What force is applied on the body by the spring when it passes through the mean position? (g = 9.8 newton/kg). Solution. There is no acceleration in the body at the mean position, hence the resultant force applied by the spring will be exactly equal to the weight of the body ie, 29.8 or 19.6 newton. Problem 31. A spring having a force constant k is divided into three equal parts. What would be the force constant for each individual part ? FE Solution. Force constant of the spring k= ©, where F is the restoring force. When the spring is divided into three parts, the displacement for the same force rediices to x/3, therefore, the force constant for each individual part F is () x13) Problem 32. How would the time period of a spring mass system change, when it is made to oscillate horizontally, and then vertically ? [Himachal 04] Solution. Time period will remain the same for both the cases, Problem 33. Alcohol in a U-tube executes S.H.M. of time period T. Now, alcohol is replaced by water up to the same height in the U-tube. What will be the effect on the time period ? Solution. The time period T remains same. This is because the period of oscillation of a liquid in a U-tube does not depend on the density of the liquid. Problem 34. There are two springs, one delicate and another stiffer one. Which spring will have a greater frequency of oscillation for a given load ? E Solution. Frequency, v = z e Force constant k is larger for the stiffer spring, so its frequency of oscillation will be greater than that of delicate spring. Problem 35. What is the ratio between the potential energy and the total energy of a particle executing S.H.M., when its displacement is half of its amplitude ? Solution, Potential energy _ Total energy Problem 36. What fraction of the total energy is kinetic when the displacement of a simple harmonic oscillator is half of its amplitude ? a) aos Problem 37. Why is restoring force necessary for a body to execute S.H.M. ? Solution. A body in S.H.M. oscillates about its mean position. At the mean position, it possesses kinetic energy because of which it moves from mean position to extreme position. Then the body can return to the mean position only if it is acted upon by a restoring force. Problem 38. What would happen to the motion of the oscillating system if the sign of the force term in the ‘equation F =~ kx is changed ? Kinetic energy Totalenersy ~ Ema a? Solution. Solution. The force will not be the restoring nature. The back and forth nature of the motion is lost. The body will continue to move in a particular sense. Problem 39. What determines the natural frequency of a body ? Solution, Natural frequency of a body depends upon () elastic properties of the material of the body and (ii) dimensions of the body. Problem 40. Why does the amplitude of an oscillating pendulum go on decreasing ? Solution. Due to frictional resistance between air and bob, the amplitude of oscillations of the pendulum gradually decreases and finally the bob stops. Problem 41. Why are army troops not allowed to march in steps while crossing a bridge? (Himachal 05] Solution. Army troops are not allowed to march in steps while crossing a bridge because itis quite likely that the frequency of the foot steps may match with the natural frequency of the bridge, and due to resonance the bridge may pick up large amplitude and break. Problem 42. A passing aeroplane sometimes causes the rattling of the windows of a house. Why ? Solution. When the frequency of the sound waves from the engine of an aeroplane matches with the natural frequency of a window, resonance takes place which causes the rattling of window. Problem 43. How can earthquakes cause disaster sometimes ? [Himachal 05C) Solution. The resonance may cause disaster during the earthquake, if the frequency of oscillations presentwithin the earth per chance coincides with the natural fre: quency of some building, which may start vibrating, with large amplitude due to resonance and may get damaged. Problem 44, Sometimes a wine glass is broken by the powerful voice of a celebrated singer. Why ? Solution. When the natural frequency of the wine glass becomes equal to that of the singer’s voice, the resulting resonance due to the powerful voice of the singer may break the glass. Problem 45. Glass windows may be broken by a far away explosion. Explain why. [Himachal 05 ; Central Schools 08] Solution. A distant explosion sends out sound waves of large amplitude in all directions. As these sound waves strike the glass windows, they set them into forced oscilla- tions. Since glass is brittle, so the glass windows break as soon as they start oscillating due to forced oscillations. Problem 46. The body of a bus begins to rattle sometimes, when the bus picks up a certain speed. Why? [Himachal 05] Solution. At a particular speed, the frequency of the engine of the bus becomes equal to the natural frequency of the body of the bus. The frame of the bus begins to vibrate strongly due to resonance. Problem 47. What will be the change in time period of a loaded spring, when taken to moon ? [Himachal 03] Solution. Time period of a loaded springy, T=2n Vk As Tis independent of g, it will not be affected when the loaded spring is taken to the moon. Short Answer Conceptual Problems Problem 1. Justify the following statements : (i) The motion of an artificial satellite around the earth cannot be taken as S.H.M. (i The time period of a simple pendulum will get doubled if its length is increased four times. [Himachal 06] Solution. (i) The motion of an artificial satellite around the earth is periodic as it repeats after a regular interval of time. But it cannot be taken as S.H.M. because it is not a to-and-fro motion about any mean position. Clearly, if the length is increased four times, the time period gets doubled, OSCILLATIONS 14.41 Problem 48. A spring of force constant k is cut into two pieces, such that one piece is double the length of the other. What is the force constant of the longer piece of the spring ? 1uT 99} Solution. Force constant, FE The length of longer part is 2x/3. So its force constant is F 3F-q 2/3 2x 2 Problem 49, In forced oscillation of a particle, the amplitude is maximum for a frequency o, of the force, while the energy is maximum for a frequency ©, of the force. What is relation between @, and, ? (A1EEE 04] Solution. Only in case of resonance, both amplitude and energy of oscillation are maximum. In the condition of resonance, 0, =o, Problem 50. The maximum velocity of a particle, executing simple harmonic motion with an amplitude of 7 mm, is 4.4 ms, What is the period of oscillation ? [AIEEE 06] Solution. v,,,, = 0A 2x22x7x10°9 = —=0.01s. 7x44 Problem 2. (i) What is meant by simple harmonic motion (S.H.M.) ? (Gi) At what points is the energy entirely kinetic and potential in S.H.M. ? ii) What is the total distance travelled by a body executing S.H.M. in a time equal to its time period, if its amplitude is A? [Delhi 09] Solution. (i) Refer to point 5 of Glimpses. (ii) The energy is entirely kinetic at mean position ie., at y=0. The energy is entirely potential at extreme positions, ie, yata, (iii) Total distance travelled in time period T =2A42A=44.14.42 PHYSICS-XI Problem 3. A simple pendulum consisting of an inextensible length {and mass 1 is oscillating in a stationary lift. The lift then accelerates upwards with a constant acceleration of 4.5 m/s* , Write expression for the time period of simple pendulum in two cases. Does the time period increase, decrease or remain the same, when lift is accelerated upwards ? [Central Schools 03} Solution, When the lift is stationary, [I T=2n/¢ Vs (ii) When the lift accelerates upwards with an accele- ration of 45m /s?, To EE Tate |— aan | Nave eeas Clearly, the time period decreases when the lift accelerates upwards, Problem 4, What is meant by restoring force ? Give one example. Solution. The force which tends to bring a vibrating body from its displaced position to the equilibrium position is called restoring force. When the bob of a simple pendulum is displaced through an 0 from the vertical a restoring force equal to mg sin Odue to gravity acts on it Problem 5. Two particles execute simple harmonic motions of the same amplitude and frequency along the same straight line. They cross one another when going in opposite directions. What is the phase difference between them when their displacements are half of their amplitudes ? Solution. The general equation for SHIM. is = Asin(at + &) As the displacement is half of the amplitude (y= 412) 50 al or sin(wt + = Asin (ot + 4) 2 ot +4 =30° or 150° As the two particles are going in opposite dire: the phase of one is 30° and that of the other 150%. Hence the phase difference between the two particles = 150~ 30 = 120° Problem 6. A simple pendulum is hung in a stationary lift and its periodic time is T. What will be the effect on its periodic time T if (® the lift goes up with uniform velocity v, (i) the lift goes up with uniform acceleration a,and (iii) the lift comes down with uniform acceleration a? Solution. (i) When the lift goes up (Fig. 14.42(a)] with uniform velocity x, tension in the string, T= mg. The value of g remains unaffected. ions, The period T remains same as that in stationary lift, [r Tease r z Te la mg me mg Fig. 14.42 (i) When the lift goes up with acceleration a {Fig. 14.42(0)], the net upward force on the bob is T mg = ma T=m(g +a) The effective value of g is (g + a) and time period is T=2n \gra Clearly, J, < Tie, time period decreases. (iii) When lift comes down with acceleration a Fig. 14.42(0)], the net downward force on the bob is mg -T' m(g -a) The effective value of ¢ becomes (g~a) and time period is Clearly, T>T Problem 7. The bob of a vibrating pendulum is made of ice. How will the time period change when the ice starts melting ? Solution. If the ice bob is of very small size, the position of its C.G. will remain same as the ice melts. Hence its time period will remain same. If the size of the ice bob is large, then ie, time period increases. 8 As ice melts, the radius r and hence the time period T will decrease. The pendulum will oscillate faster. Problem 8. The amplitude of a simple harmonic oscillator is doubled. How does this affect (i) periodic time, (if) maximum velocity, (ii) maximum acceleration and (io) maximum energy ? [Chandigarh 03)Solution. 1 ation per unit displacement Toe a Asthe acceleration per unit displacement isa constant quantity, Tis not affected on changing the amplitude, Gi) May = OA When amplitude is doubled, maximum velocity is also doubled. (il) yyy = RA When amplitude is doubled, the maximum acceleration is also doubled. E=2n? mv? A? ie, Ex AP When amplitude is doubled, the energy of the oscillator becomes four times (io) Problem 9. You have a light spring, a metre scale and a known mass. How will you find the time period of ‘oscillation of mass without the use of a clock ? Solution. Suspend the known mass m form the spring and note the extension /of the spring with the metre scale. Ifkis the force constant of the spring, then in equilibrium m_t 2m ‘i 8 M=mg or is ar, period T ‘Time period of the loaded spring, T=2n & So by knowing the value of extension I, ti can be determined. Problem 10. A man is standing on a platform which oscillates up and down simple harmonically. How will the weight of the man change as recorded by a weighing machine on the platform ? Solution. As the platform moves from the mean position to the upper extreme position or from upper extreme position to mean position, the acceleration of the oscillating system acts vertically downwards and hence ‘weight of the man will decrease. On the other hand, as the platform moves from mean position to lower extreme position and then back to mean position, the acceleration acts vertically upwards, Hence weight of the man increases, Problem 11. The frequency of oscillations of a mass m suspended by a spring is v,. If the length of the spring is cut to one-half, the same mass oscillates wit frequency v2. Determine the value of v,/v,- [Chandigarh 03] Solution. Let kbe the force constant of the full spring, Then frequency of oscillation of mass m will be at fe 2n When the spring is cut to one-half of its length, its force constant is doubled (2k). OSCILLATIONS 14.43 Frequency of oscillation of mass will be 1 [ek Vn valy, =v2 Problem 12. All trigonometric functions are periodic, but only sine or cosine functions are used to define SM. Why ? [Central Schools 03} Solution. All trigonometric functions are periodic. ‘The sine and cosine functions can take value between —1 and + Lonly. So they can be used to represent a bounded motion like S.H.M. But the functions such as tangent, cotangent, secant and cosecant can take value between 0 and ® (both positive and negative). So those functions cannot be used to represent bounded motion like S.H.M. Problem 13. A simple harmonic motion is repre- 2 sented by o + 0x =0, What is its time period ? . P, IATEEE 05] Solution. Cleary, > =—ar or a x fz ms Tare Fam (z=. a Nae Problem 14. Does the function y a periodic or a simple harmonic motion ? What is the period of the motion ? [AIEEE 05] Solution. Displacement, y= 8 <2 sin at xcos ot xo at Time period, jin? ot represent Velocity, As the acceleration a is not proportional to displace- ment y, the given function does not represent SHM. It represents a periodic motion of angular frequency 2a Time period, ca oe ‘Angular frequency 20 0° Problem 15. The length of a simple pendulum executing SHM is increased by 21%. What is the percentage increase in the time period of the pendulum of increased length. [AIEEE 03) Solution. Time period, rex ft ia, Tall? 8 ‘The percentage increase in time period is given by aT 1a AF <100= = «100 T 2114.44 PHYSICS-xI HOTS Problem 1. Two simple harmonic nrotions are represented by the equations : xy oS sin Qnt-+ a/4), xy =SV2 (sinQnt + cos2nt) What is the ratio of their amplitudes ? {Roorkee 961 Sol sin 2at +x/4) A a H V2 (sin 2nt + cos 2nt) =10 sin (sin 2nt cos m/4 + cos2x t sin n/4) Problem 2. The bob of a simple pendulum is a hollow sphere filled with water. How will the period of oscillation change if the water begins to drain out of the hollow sphere {from a fine hole at its bottom 2 Or The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscilalting body gets suddenly unplugged. How would the time period of oscillation of the pendulum change, till water is coming out ? [AIEEE 051 Solution. Time period, T an 8 As water flows out of the sphere, the time period first increases and then decreases. Initially when the sphere is completely filled with water, its C.G. lies at its centre. As water flows out, the C.G. begins to shift below the centre of the sphere. The effective length of the pendulum increases and hence its time period increases. When the sphere becomes more than half empty, its CG. begins to rise up. The effective length of the pendulum increases and time period T decreases. When the entire water is drained out of the sphere, the CG. is once again shifted to centre of the sphere and the time period T attains its initial value. Problem 3. The period of vibration of a mass m suspended by a spring is T. The spring is cut into m equal parts and the body is again suspended by one of the pieces. Find the time period of oscillation of the mass. [AIEEE 02] Solution. The force constant is inversely propor- tional to the length. If k is the force constant of the original spring, then the force constant of each part will, be nk. 7 Pa vin Hence Problem 4. Two simple harmonic motions are repre- sented by the equations 4 =041 sin (100 nt + x/3)and_ yp =0.1 cos nt What isthe phase difference ofthe velocity of the particle 1 with respect to the velocity of particle 2? {ATEEE 05] Solution. Velocity of particle 1, Ea =0.1 cos (1008 + 2/3) 100 x =10 n c0s (100 xt + r/3) Velocity of particle 2, =th. ade = 0.1 c0s (nt + 2/2) Phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is A9=4-t Problem 5. A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency Gy, An external force, F(t) « cos at (oF) és applied to the oscillator. How does the time displacement of oscillator vary ? IATEEE 04] Solution. With natural angular frequency @, the O.1(-sin at)x x =—0.1 nsin nt RK on acceleration of the particle at displacement y is %=- a5 Y The external force F(t) «cos ot has an angular frequency « The acceleration produced by this force at displacement y is d=ay ‘The net acceleration of the particle at displacement yis anata =— bys af y=—(eR—oP)y ‘The resultant force on the particle at displacement y is F=ma=-m(ug-«?)y or mE Clearly, ye (a —@?)The point of suspension is now moved uproard according to the relation y=Kt? (K=1ms), where y is the vertical displacement. The time period now becomes T, What is the Problem 6. A simple pendulum has time period T, ratio T3/T} ? Given ¢=10 ms TUT 05] Solution. In first case, ol) In second case, displacement y = Upward velocity,» a =2Kt Upward acceleration, a=2 K=2 x Ims?=2 ms T 7 2n oo "eee ge2 7 Problem 7. The bob of simple pendulum executes SHM in water with a period t, while the period of escillation of the bob is ty in air. Neglecting frictional force of water and given that the density of the bob is 42% kgm, find the kg ni relationship between t and ty ? AIEEE 04] Solution. In air, tg =2x E 8 Let V be the volume of the bob. Then Apparent weight of bob in water ‘Weight of bob in air — Upthrust Vps'= Veg - Vog (rt Density of bob, = gm? Density of water, «= 1000 kg m= 1000%3)_g : eae = ® ( 4000 Js 4 Time period of the pendulum in water, a x Pantone Problem 8. A mass M is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes ST/3. What is the ratio m/M ? IAIEEE 03], t 2ty. OSCILLATIONS 14.45 Solution. With mass M, the time period of the spring is mM re2n [M Ve With mass M+ 1 the time period becomes x {@ m =2n Sate fiom a or 3 a-Mem or Mam . mat, M9 Problem 9. Two bodies Mand N of equal masses are suspended from tto0 separate massless springs of spring cons- tants k, and k, respectively. Ifthe two bodies escillate vertically, such that their maximum velocities are equal, then find the ratio of the amplitude of M to that of N. IATEEE 03] Solution. The maximum velocity of body in SHM is given by 2, 1 Aon ale Giver Pag, (M)= Pay (N) Fr A, {2 ae or aye ay elim my = or Lem Problem 10. A particle at the end of a spring executes simple harmonic motion with a period t,, while the corresponding period for another spring is ty. What is the period of oscillation when the two springs are connected in series ? AIEEE 04] Solution. If a force F applied to the series combination produces displacements t, and t, in the two springs, then Fa-kyxy=- kyr, and x)= Total extension, oe or = Inky ky, +k, -. Force constant of the series combination, pa ttS14.46 PHYSICS. XI Period of oscillation for the series combination, roan | vay [MOD 3 [im kT kk Vk," ky w meal ez) -bele) “e) o Patee Problem 11. A particle executes simple harmonic motion between x= ~ Aand x =+ A. The time taken for it to 0 from 0 to A/2 is T, and to go from A/2 to A is T,, Then how are T, and T, related ? IT Screening 01) Solution. The displacement equation for S.H.M. is Asin ot AU YS ALD Acasinot, of Lesinat, 2 2 or of, =* or ae t=, +Ty At x=A A= Asin o(T, +1) or 1=sin o(T, +7) x ® T,+T)-2 or 1 +h-= or o(%+%)=5 a5 non t ~2 = ear, 20 20 60 30} Problem 12°Bweo simple harmonic motions are represented by the equations y= 10 sin (12 {4 1), yp=5 (sin3nt+ V3 cos3nt) Find the ratio of their amplitudes, What are time periods of the two motions ? UIT 86; MNREC 90] Solution. y, 10sin F(2t+1) x =10sin(3nt+™ oll (9) ” Ys = 5 (sin 3x t + V3 cos3xt) 7) (were sensed =10( sind cos $+ sant sin) 10 t+ a or sin(3n 2) 0) ‘The general equation for SHM is Y= Asin (at + )= Asin (2214 4) (3) Comparing equations (1) and (2) with (3), we get 2n Problem 13. A point particle of mass 0.1 kg is executing SHM of amplitude 0.1 m. When the particle passes through the mean position, its kinetic energy is 8x 10-? J. Obtain the equation of motion of the particle ifthe initial phase of oscillation is 45° [Roorkee 91] Solution. Here m=O1kg, A=O1m, E=8« 10-7} =45° =" rad 4% ry K.E. at the mean position =(E,)max pmo a sx 10"? =2 sor a? x 01)? or o?=16 or @=4radst The equation of motion for the particle is y= Asin of =0.1 sin (4 t+/4). Problem 14. A simple harmonic motion has an amplitude A and time period T. What is the time taken to travel from x= Atox= A/2? [REC 92] Solution. Displacement from mean position A_A =a-A.A 2 2 When the motion starts from the positive extreme position, Problem 15. A block is resting on a piston which is moving vertically with simple harmonic motion of period 1.0 second. At what amplitude of motion will the block and piston separate ? What is the maximum velocity of the piston at this amplitude ? [Roorkee 85] Solution. The block and piston will just separate when Mmax=8 OF go A= ant Maximum velocity of the block, 2n 2% 3.142 10 Umax =OA=— A 0.248 = 1.56 ms’Problem 16. A block is kept on a horizontal table. The table is undergoing simple harmonic motion of frequestey 3 Hz in a horizontal plane. The coefficient of static friction between the block and the table surface is 0.72. Find the maxinuum amplitude of the table at which the block does not slip on the surface. Take g= 10 ms". [Roorkee 96] Solution, Maximum acceleration of the block, Aya, = OPA :, Maximum force on the block, F = mia... =m a" A Frictional force on the block =}. mg The block will not slip on the surface of the table if mo? A=p mg 2. Amplitude, ug _072%10 wo nv Problem 17. Springs ofspring constants k, k, 4k, 8k, ‘are connected in series. A mass m kg is attached to the lower end of the last spring and the system is allowed to vibrate. What is the time period of oscillations ? Given m= 40 g and k= 2.0 Nom Solution. Here m= 40 g =004 kg, k=20N m=! =20x100 Nm" * ‘The effective spring constant k of the series combination is given by a] ae 8 i 1a tate teoeae + Kk” 2k” 4k” 8k 1 7 9 ==-|—|= f finite GP. i[fAal i or K'=k/2 rere {ote kK « 2x2. 7 = 0.1265. Problem 18. A uniform spring whose unstretched length is I has a force constant k. The spring is cut into two pieces of unstretched lengths I, and l,, where l,= nl, and nis ‘an integer. What are the corresponding force constants k, ‘and k, in terms of n and k ? What is the ratio ky /k, ? Solution. Here! +h, and =n, or hen As k= OSCILLATIONS 14.47 Problem 19. A horizontal spring block system of mass M executes simple harmonic motion. When the block is passing through its equilibrium position, an object of mass m is put on it and the two move together. Find the new amplitude and frequency of vibration. [Roorkee 88] Solution. Original frequency, 1 {z 2nVM Let A= Initial amplitude of oscillation v =Velocity of mass Mwhen passing through mean position Maximum K.E, =Total energy 2 ao? = ka? 2 z= k M When mass m is put on the system, total mass =(M-+ m). If vis the velocity of the combination in equilibrium position, then by the conservation of linear momentum, v Mo Mv =(M+ mv = f=(M+m)o or v= If A’ is the new amplitude, then 1 2lyar > vt =i kA 5 (M+ mv => o | A= em Mem Mv k k Mem _ [Mem oM Vk" Mem E fe M+m i [e 2n\M+m Problem 20. The bob of pendulum of length | is pulled aside from its equilibrium position through an angle ® and then released. Find the speed v with which the bob passes through the equilibrium position. (Kurukshetra CEE 96] New frequency, Vv14.48 PHYSICS-xI Solution. The situation is shown in Fig, 14.43 Fig. 14.43, — to NCERT Exercises 14.1. Which of the following examples represent periodic motion ? (8) A swimmer completing one (return) trip from one bank of a river to the other and back. (ii) A freely suspended bar magnet displaced from its N ~ § direction and released. (iii) A hydrogen molecule rotating about its center of mass. (iv) An arrow released from a bow. (v) Halley's comet, Ans. (i) Not periodic. Because the motion of the swimmer is not repeated over and over again after any fixed time interval. (ii) Periodic. As the magnet is released from its displaced position, it oscillates about the N-S$ direction with a definite time period (iii) Periodic. The motion of the hydrogen molecule rotating about its centre of mass repeats after a fixed time interval. (iv) Not periodic, The motion of the arrow does not repeat itself after a fixed time interval. (0) Periodic. Halley’s comet appears after every 76 years. 14.2. Which of the following examples represent (nearly) simple harmonic motion and which represent periodic but not simple harmonic motion ? () The rotation of earth about its axis. (i) Motion of an oscillating mercury column in a U-tube (tii) Motion of a ball bearing inside a smooth curved bowl, ‘when released from a point slightly above the lower ‘most point. (iv) General vibrations of a polyatomic molecule about its equilibrium position. Ans. (i) Periodic but not simple harmonic. The motion of the earth about its axis repeats after every 24 hours but it is not a to and fro motion. Clearly, hh =OB=05- BS= =1(1~ cos ) Let vand v’ be the velocities of the bob at position and A respectively. Then by the conservation of energy, © ney? + mgh or v= Jv? -2gh =v? ~2g! (1-cos 0) ii) Simple harmonic. The restoring force is propor- tional to the displacement of the mercury column from the equilibrium level (ii) Simple harmonic. The motion of the ball bearing is to and fro about the lower most point and the restoring force is proportional to its displacement from that point. (io) Periodic but not simple karmonic. A polyatomic molecule has a number of natural frequencies. In general, its vibration is a superposition of SHM’s of a number of different frequencies. This superposition is periodic but not simple harmonic. 1433. Fig. 14.44 depicts four x-t plots for linear motion of a particle. Which ofthe plots represent periodic motion ? What is the period of motion (in case of periodic motion) ? Be Fig. 14.44Ans. (i) Plot (a) does not represent periodic motion because the motion is not repeated after a fixed interval (i) Plot (b) represents periodic motion with T = 2s. (ii) Plot (©) does not represent periodic motion. The repetition of merely one position is not enough for the motion to be periodic. The entire motion during one period must be repeated successively. (iv) Plot (@) represents periodic motion with T = 2s. 144. Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, ancl (6) non-periodic motion ? Give period for each case of periodic motion fois any positive constant } () snot - cs of (iii) 3 cos (n/ 4 —2ot) (0) exp(- ot?) Ans. (ii) si oot (iv) cos wt + ens 3ot + cos Sot (vi) 14 of + 0. (i) Here x (t) = sin wt ~c0s ot = V2 (sin ct cos x/4—cos of sin / 4) = ¥2sin (wt ~ /4) Moreover, x(t +2n/o) =V2sin [a(t + 2/0) ~x/ 4) = ¥2sin (et + 2n- 1/4) = V2sin (ot - x/4)= x(t) Hence the given function represents a simple harmonic ‘motion, with T = 2n/ and phase angle =~ x/ 40° 7/4. (ip r(t)=sin? ot =] sin at sin Sor) [+ sin 30 =3 sin @ — 4 sin? 6] It represents two separate simple harmonic motions but their combination does not represent SHM. Period of 2 sin ot =2© <7 4 © Period of } sin 3ut = 2% = E 4 30° 3 ‘Thus the minimum time after which the combined function repeats is T = 2x / a Hence the given function is periodic but not simple harmonic. (iil) Here x (t)=3.c0s (x/ 4-201) = 300s [-(2ot ~ x/4)} = 3.05 (20t - x/ 4) [> cos (~ 8) = cos 6] It reptisents SM. with period T= 2% =. 20 @ (iv) x (2) =c0s at + cos Su! + cos Sot cos ant represents S.H.M. with period = 2% ° 2 cos Se represents S.H.M. with period = cos Sat represents S.H.M. with period OSCILLATIONS 14.49 ‘The minimum time after which the combined function ropeats its value is T. The given function is periodic but not simple harmonic (©) x(t) -exp(- 0%? 1 is an exponential function. It decreases mono- tonically to zero ast > ~, Itnever repeats its value. Ibis a non-periodic function. (i) x(t)=14 ot + 0%? ‘Ast increases, x ()inereases monotonically. Again, as t+ «, x(t)» =, The function never repeats its value. So x(t) is non-periodic. 445. A particle is in linear simple harmonic motion between two points, A and B, 10 cm apart. Take the direction {from A to Bas the positive direction and give the signs of ‘velocity, acceleration and force on the particle when itis (a)at the end A, (bat the end B (© at the mid-point of AB going towards A, (a) at 2 cm away from B going towards A. (€)at 3.cm away from A going towards Band (fat 4.cm away from A going towards A. Ans. Positive direction > Zer0 a BE O Co f—3em Eee haem an |. 14.45 Position | Velocity | Acceleration | Force (ALA | O(atextreme [+ ve facts | + ve (acts position) | from A to 0) | from A toO) (Ate — | Ofatertreme |— ve (acts | ~ ve (acts position) | from B 10.0) | from B to O) (at =veand — |O(atmid- | 0 (at mid- midpoint 0, | maximum | point) point) going {acts from 0 towards A | to A) Marc, — |-velats | -velacts | -ve (acts going from Cto0) | fromCto0) | fromC 00) towards A Garo, — |svetads | +vetacts | + ve (acts going from D100) | from D t0 0) | from D to 0) towards B (ALE, — [=vefacts | + ve acts | + ve (acts going from E to A) | from E to.) | from E to.0) towards A | 14.6. Which of the following relationships between the ‘acceleration a and the displacement x of a particle involve simple harmonic motion ? (a =07x (b)a =-200x? (ca =—10x (d)a = 100%, ‘Ans. Only (c) represents S.H.M. because here a «x and a acts in the opposition direction of x.14,50 PHYSICS-XI 147. (a) A particle in SHM is described by the displacement function, 2n 1) = Acos(ot =2, x(t) (ot +) o=F If the initial (¢ = 0) position of the particle is 1m and its initial velocity is m oms™?, what are its amplitude and initial phase angle ? The angular frequency of the particle is xs~' (8) A particle in SHM is described by the displacement function, Qn - If the initial (t = 0) position of the particle is 1 cm and its initial velocity is xm s~', what are its amplitude and initial phase angle ? The angular frequency of the particle is xs} . Ans. (a@) At f=0, x=1cm and v=nems"!. Also, 1 Xx ()= Bsin(ot +a) @: o=ns In SHM, displacement at any time f is given by X= Acos(ot + §) 0. x=1 therefore 1 Acos(ox0+ 4) Acos ¢=1 Now velocity, Since, at or Ai) [Acos (ot + 4)] a = ~Aosin (ot +4) Q, v=nems”, so we have n= — A(n)sin (@x0+ 6) Asin $= -1 Squaring and adding equations (i) and (i), we get A? cos* 6+ A? sin? =F + (1)? A? (cos? $+ sin? §)=2. or A? (1)=2 2 V2cm. Dividing equation (i) by (), we get at 7 or tang Again at ¢ or (it) or 3x "4 (b) At t=O x=1cmand v=nems™ Also, ‘a Given x= Bsin (ot + a) Since, at t = 0 x= 1 therefore Bsin (ox0+a) Bsina=1 Now velocity, v= 8 4 (Basin (ot +a)) = Bacos (wt + a) In or or i) Again, att =Q. v= ems", so we have B(x)cos (@ x0+ a) Boos a =1 oii) Squaring and adding equations (i) and (ii), we get B sin? a + B’cos* a = + Por B v2 em. Dividing equation (i) by (ii), we get or or 14.8. A spring balance has a scale that reads from 0t0 50 kg. The length of the scale is 20 om A body suspended from this spring, when displaced and released, oscillates with a period of 0.60 5. What is the weight of the body ? ‘Ans. The 20 cm length of the scale reads upto 50 kg, s0 F=mg=50%98N,° y=20cm=020m Force constant, k= £ = 5098 y 020 ‘Suppose the spring oscillates with time period of 0.60 s when loaded with a mass of M kg. Then iM i 2M 2450 Nm~? or T? 4 m= Dk _ (0.60)? «2450 4x? 4x (3.147 Weight = Mg = 22.36 x9.8 = 219.13 N. 14.9. A spring of force constant 1200 Nm=" is mounted horizontally on a horizontal table. A mass of 3.0 kg is attached to the free end of the spring, pulled sideways to a distance of 2.0 cm and released. (i) What is the frequency of oscillation of the mass ? (ii) What is the maximum acceleration of the mass ? (iti) What is the maximum speed of the mass ? Fig. 14.46 Ans. Here k= 1200Nm"!, m=3.0kg, A=20cm =2.0x10-? m 22.36 kg (i) Frequency of oscillation of the mass, 1 fk, 1 [i200 2nVin 2x314V 30 x20=32 2x34 (i) Angular frequency, [E [9200 ay gt Vaan