10 SEH OXETAG yy unot! res | MNCHANTES cent, At OL ATTN go 0 te nn forwe Fema 02 NOS Mayne sea eae ace Pane fre a ce AHH eee emyor E01 : =o mcsesyeee’ ak to INO HEFO PODiLlog Oe ah te ab MC, ex wit reat Benes, the pointer - ygxte cae In) In thiscaise, P02 . Oe Dy we = Zemvor = 20 mT * ea alto the 2er0 OF the teate, so nm forwsr Hence, the pointer will read 20 ry at rest is dred fr01 HeHRC Of 2 yy, EX & A Ano te Fr ao Nit, Calera the may mn int will be ca" Sol. Loss in gravitational potential oneTHy = Gain in potential energy by the Spring A f | Here, m= 1k, ¢=9-8nv/s and C =490Nm. Ix98Q+y or 10027 =4x48 or 2x? -x-2 =0 whence xo teem Poe $.10. Potential Energy Curve. If the potential energy U of a ‘cle i i iti ia y Particle is a function of position (ie changes nom Point (0 point), then a graph may be plotted to ‘ton variation of Potential energy with the position of the particle, Such a Brot 7 pow &8 potential energy curve.* In case, if the particle is cnet U wl be the function of on) long X-axis, the pte will of x-coordinate onl id e fi ima 3 ¢ , and then the force « ‘ve in nature) on the particle will be given by a ) Fig. 5-12 sh aT ae @ ig. 5-12 shows il a motion, The slope (IU Potential energy curve for one dimension@! (F=~dU/de), acting on OF this curve at any point gives the for hoa a © Particle placed at that point. The slop? 8 the equation of its motionCONSERVATION OF ENERGY | 199 a) is positive, whenever U increases with th # and CD of the curve) and ney © ine (eg. for the part GA : the force ~dU/as) is negative (i, a “galive X-axis) or directed to t lef whenever the potent ia ‘asing with x, and Positive or ected to the right Whenever energy is decreasing with x ig. 5.12). This means that the force, acting on the panicte at any point, es fo bring the panicle to the von of lower potential energy y In the Fig. 5.12, at the Points 1 B,C and D, the potential ergy U is” minimum or aximun, the value of dU/dx is tro. Therefore, if a particle ig such a point with zero it will experience no —dU/dx ibrium, . In the figure, B and D are the Fig 5.12 ints corresponding to maximum Potential energy. A. Particle at rest at uch a point will remain at rest. However, ifthe particle is displaced even he slightest distance from this Point the force Foy = -dU/ax will tend 10 push the particle farther away from the equilibrium Position. Hence the points B and D are the positions of unstable equilibrium, In the figure, 4 and C are the Points corresponding to minimum potential energy, A Particle at rest at such @ point will remain at test. If the Partic F ‘onstant in a region, the force (F=—dU/dr), acting ona pa region, is zero. In such a region, if a Particle is displaced from one point to the other, it will remain there without experiencing any restoring force, So we call the region of constant potential energy as the region of neutral 'quilibrium. For example, a book placed on a table anywhere remains in ‘quilibrium, Potential well. Now, let us discuss about the point on near to it, which is of great lergy Up. The total energyE = K + Uof the Particle has been indicated200 | MECHANICS 7 tant during the motio, by a horizontal line (PQ) and will remain COMET On oe ry a eee ‘ vause the system iS OTE the curve and s0 the if the the point A, because | in by the ordinate ©! dding to the total netic ys eee ‘The , horizontal line correspont ne particle ie cnet energy K =E — U. The dQ. When at Pi = interses Ix curve at P an is in the form of at a intersects the U-x eal the total energy E is a nt _Potentig, ‘inctic energy is zero al ‘cle is acted by @ restoring foro cao aoe Position, the Parr particle to move towards A ang fs rease in potential cng, potential energy decreases gradually. This a article, because appears in the form of the kinctic ee the point A’ the part ie total energy should remain constant. d maxi mile " ul ‘al energy (Up) and maximum kinetig possesses the minimum potential Ms icle moves tows energy. Now, after crossing the position A the partic! form, th ards Q and when at Q its total energy becomes in ie poten a © Particle acquires rest position and again the particle is moved towards A by the restoring force. Thus, the particle oscillates with a certain Period 7 between the tuming points P and Q about the point A and is confined in the region between these points. Such a region is called bounded region or potential well and if the particle is limited to this region, it is said to be in bound state. Hence the bounded region is present near a point of stable equilibrium. The difference of potential energy corresponding to the top and the bottom of the potential well is the binding energy for that potential well. If a particle is in a potential well (or sink), it will remain there in the bounded staie in the condition that its total energy E is less than the potential energy corresponding to the top point (e.g. B). The potential energy U=f(x) can be expanded in a Taylor’s series about the point of stable equilibrium Xp iz. dU PU) (x - x0)? YX =v (te) (x — x0) + a ° ag G2 Ixy 2 3, 3 +{¢U & = 30) +. xo ! where U,, (a ty ete. are the values at x =o As the point A is the oint of le equilibri ; aa: point of stable equilibrium (ie minimum potential energy) the value ofCONSER) 'VATIOS N OF ENE; ERGY | 204 zero and (<3) should shou ghould be di }xy Pe postive quantiy : 1 say C, . dU) Fe ceinomeot (28) tig ti Ceol Ge Jag 0 Bet t U =U t+ § 09241 a 2! 31 Oa) Xe sume the origin at x9 and the minimum ¢, ¢ oe for @~ 20) We have SnerBy Up =0, then ii E wil ¢ (26 ; The force acting on the particle is given by ) H qwW__¢_ ax . .. (20) . ete. are zero, then =- at F Ce and U= 4c? a a) acase, if a graph is plotted between U and x, it will represent a bola ie, the potential well is parabolic in ‘shape. Here, the force acting on the particle is proportional to the displacement and the positive constant C is called the force constant. Actually, this is the case of a harmonic oscillator, for which essentially a parabolic potential function should exist. The case of the harmonic oscillator with a few typical examples will be dealt in detail in chapter 9. In case of the oscillations of small amplitude, the displacement x is small so that higher powers of x are negligible. Thus, from eq. (26) and (27), we have =40? and F= —Cx. Hence, for small amplitude oscillations the potent tial well may be treated as parabolic and the oscillations are simple harmonic in type. IEC), Cy, ... are not zero, the potential tnergy well will not be parabolic and in seneral the oscillations of the system will be : harmonic. Fig, 5.13 fag Figure below shows the poten iagra Daun lect, potential energy U being pploted against inter ru “evalues of the following : OF quitibrium internuclear distaiice ro - Cy," Insuch ial energy diagram of a gst I?sor | SMCHANHCS a Titer ee ett ai ea (ure od nk Ltt yay { \ t ro cae TE eo a Gi " webs. +4—F I LI 08 ) Near the hoo of = | | | te wo potential well, it is 4 Fig. 5.14 we and hence the wo al energy is given by 7 v 21d, whence C #207 z andx =4AU, then, gevie, U = 20V Taking U between ~4eV and — is 6 *10— 2 16x 1 GX age o 2 {AU = — sa eerir') caBEp = 16evin 0792 ® ¥ EE d U=Up ~ Px + Oe Fin Id ential energy fie ty Ex. 2. A particle moves in ae Wjorce constant At what point does ihe on expression for the force. Calculate fo venshes sth point of sable equilibria : Sol. F=-Gb = +P - 20% Evidently, this is a linear restoring force and the force constant is 29, ‘The force vanishes, where we Oie., P-2Qx=0 x= P/2Q. (2, Now, £4 ~ 429 ie, if Q is positive, atx = P/2Q, there is the minima y ae potential energy curve and hence this point is in stable equilibrium. * 5.11. Non-Conservative Forces. A force is said to be non-conservative, if the work done by that force on Particle that moves between two points depends on the path taken betwen those points. Frictional and viscous forces are the examples of non-conservalitt forces. In case of such forces, if a particle moves from A to B and theo from B to A, the work done around the closed path is not zero, but tht particle itself loses its kinetic energy along both paths, For example, if push a block over a rough table between any two points by various pathsCoven, wee oe Pe K+U~ Ew Coenen ial ax+ au oO = E fs vrome By cOREETUATIVE FAME Fe oe a yor! Vo = ~A& TAB) ng, © pol © weatia) NOTES 14> We te, PUI he Be ee yppose, that ip eddinion ma ey Net ree: ve tO FrAcTION acts or he a ee f nd ; ce py the frictional force and W bers wor 08 We) am the parle shows be o ‘k ae yinctic ENOTES of the particie, Ths ex Wy + We = SK We =-al, Bot AK+AU =W3 ation shows that if a frictional force aces on» oy em pris 04” ‘cal energy is not constant, bat changes by otal Oey the frictional forcs. Eq. (29) can be arrives as eee AE=E-Ey is the work done dy fi Wy, which is the a ois Shaan Bot Wp er ve see from eg. (30) that the final mechan ie v ws) work men pi Jess than the initial mechaniosl energy £) (= XK; Bere mechanical energy appears im the form of heat. Tee eat ora Toe foped is exactly equal to the mechanical caczgr Gissipsted, Thc reed ory produced, is equal to the work dome dy the particle, 10k Ge oe Bhergy developed, then evidently Wy = —O aad hemes we get AE+Q=0 ee is nO he of the mechan ant This shows that there peat eDergy of the system so far conservative and Frctimal foros are AO one gstem. Therefor ioe i ee may agsin de aid tee amend Similarly if other non-conservative forces are &SAE, sdoa the mechan ed into other forms of onengy, Uherohore, Dae kata energy energy lost is convert ingmeral is conserved. 1.€., AE + Q + change in other forms of energy =O @ Thus, the total energy is always conserved while kinetic aest prxcatial mergies are conserved only when comservaline forces act 1s $12. Motion under Friction. ‘When one solid body is made to slide over the surface of anceber, & resisting force comes into play in between the two surfaces. This ress force tends to destroy the relative motion betwees the (we baxtees and is eee of friction, Even when there is no relative moms derweest es, frictional forces may exist between the surfaces of coatasti | vel, 189 very COMPIICar. ic tev on an atomic fe Yel -apie HANtcs croc a e Cc} viewed 4 Wy mata! are iff os com 10 cling togcthe, c fom ~ 4 , ° are the AON apoms SAP APAH aang son, when Friction, WRT gurfaces ON fcontal Yigg alone |. friction Was thought 1, mercly full of of this ere d then, a8 - y the yrfaces at, eration ensues: FOr = Sieve the sliding body ove, vas be : : eter be very simple I ition originated here is 0 loss of encray in and the fi ye, because ed when one body aoa | MECT phenomenon : ints of enovare many Ponts) There My Piding irregularities and ¢ i" . eum ¢ : but this canne ‘ er is a © dy the bumP® We know that actually Pp of power Joss 1s that as the slider | that process ther. The mechanisne ond then produce Waves and | slides over the ole" 5 deform . + the bumps Ti bodies. f Snaps over the Bumps, INEPT in the two + catoee, Tat atomic motions an het ise sy rests Of horizontal . The Suppose that a bo ier the reaction R due to the surface - ide +R upward direction and is equal yin : 1 | weight of the body W. On ee ‘ : 1 the plane, appuicn | i small force parallel to , cause at FORCE OFF body not in general move, the surface of separation a force equal FRICTION FORCE and opposite to the applied force i) comes into existence. This is the force w | of friction. On increasing the applied Fig. 5.15 1 force, the force of friction increases = i upto a limit. At this instant the body is just about to move and the force of | friction is just enough to preserve equilibrium. This maximum frictional {force is called limiting friction. On further increasing the applied force, | imentally that for any pair of two the body starts to move. It is found experii srufaces, the limiting friction F bears a constant ratio to the normal ees eet i oa be ratio i called the static coefficient of iply coefficient of friction for the given pai i denoted by the letter .. Thus, aiveo pai of surfaces aad * ye = imiting friction _ F | -, ; normal reaction — R* iadepeutest ot the toon or ten iy Pairs of substances, bu i is provided that itis mot excessive nt Contaet and of the weight W, Friction plays a very imy ‘ i on portant . friction of the ground it would be ae ao everyday life. Thus, without possible for us to walk, to fix a nail into a block, to climb tres ii applied to moving heck thee ie Knol etc. It is the friction ‘of the brake In many Cases, we find a , that frict machines must be reduced to stig Petween the moving parts of the a minimu it i m and yet it is necessary even forL hem. This is done by t 1: bricalion reducing the area o¢ ; . meee jan Upht c124. Dynamic or Sliding Evtetiom al etna ead f We have seen above that then ty with the applied force and attaing nt ition : te pash the body gently so they Matlin or tgs this oo without acceleration, then the retanta seamed ine Pa sha. sliding friction. This is measured tan force in the fam ction slide maintain the body in uniform mallee APNE Coreg p dynamte we Me nal n R (equa On The ratio of ¢ Fequived arvmte Teton. Te vale a alle the om 8h ee value is tess than the st HEAL OF atldinng ce which means that a less force is required sg ne Mica ut feition velocity than is required to start the body sliding Vitti a constant sliding, 1! constant at moderate velocities, the dynamic friethon | 5.12-2. Angle of Friction, =F anterior a of li i sey he maximun into play to maintain equilibrium), the angle makes with the common normal is called the w of friction. | Let us consider a body of load W, placed on the surface of another in limiting equilibrium. In this condition, the normal reaction R of the first body acts normally along OR. If the motion of the body tends to take place along OF, then the force if limiting friction eR (in equilibrium) will act as amount of friction ix called which the resultant reaction Sot indicated in fig. 5-16. Now, the resultant reaction y is nd is oe PHVR +R = RVI and the anlge which this resultant makes with normal is the angle of Snction A, i. itis tana = 4 = word = tan! ty i . 6 i Now, taking O as vertex, the normal OR as axis and Led. Hed bo | ‘viction 2 as the semi-vertical angle, draw a cone, then (his cone nail cone of friction. rok’ | S13, Inclined Plane. faa lane, W can be resolved oe Ifa body of weight W rests on an inclined plane, Waosd is qual tO the into two components W cos @ and W sind. The free el 1 fo the normal reaction Rand W sind is parallel £0 the PAE TN icing «Te norn é iontal, the position ‘ndlination @ of the plane to the horizon! aepoint of sliding. In this cond Wind = aR = 4 Weos@ or But wo tand where A is angle of frieti a 0 a of OM 7 ater of slidin ‘Thus, the body is on the point o jane is equal (0 when the inclination of the pI the angle of frinetion, ; 8 F ill If the force Wsin O>4R, the body slide. Now, if the mass of the body ae is Mgsin@—pR » plane 6 re! se parallel to the P cea Me no eae Faas the acceleration cof the body on the inclineg plane, Mgsin 0 ~ 4 Mg.cos 0 = M = g (sin — #6088). Fig. 5.17 is given by $.14, Rolling F Rolling friction is the resistance to motion when one surface rolls on the other, eg, rolling of the wheels of a cart on a rod or a railway carriage on stcel rails, Rolling friction, of course, 1s much less than sliding friction for surfaces of the same material. That is why vehicles are provided with wheels. In rolling, an appreciable amount of slipping or sliding of one surface over the other occurs and the frictional resistance to this slipping or sliding, is the cause of rolling friction. 5,15, Stability due to Friction. When a body, say in the form of a block of wood, rests on a horizontal stable equilibrium because its weight , plane surface, it remains acting vertically downwards at the centre of gravity G, is balanced by the equal and opposite reaction R acting there. Now, if a horizontal force f is applied at D, it does not slide along the plane so long as f Cp, th the body will Fig. 5.20 Findricat body atthe of the cy Se rope aC. element PQvot th jon of nds of thehe rope along ¢ r fag be a difference in ten force, thet emt, Lt wir thee oh ; ee g = p dO. eh rapt oof a top T ee 7 a0, loge T1 =k Oe ey fogs 7771 = Mo a: jog. Ts ariation of tension T with the any on in the rope at E is 72, then n represents the wle GOH is ¢@ and the tensic Ty/T = e*. 1 is coiled several times round the cylindircal bod doy Sees T/T; may be so great that a small te 4 aA can hold a very heavy weight attached to E. appli $.17. Band Brake, principle, discussed above, is employed to consturct many fi kes. These brakes are used to measure »wer of machines and among which the 4 one is band brake. It consists of a fixed on the rotating shaft of the ne whose power is to be determined, 4 tend of a cord passes over a pulley once, twice or thrice etc. and its two ends are cied to spring balances A and B, ded Irom a rigid support (Fig. 5-21). Now, if T; and Tz are the readings of the two we balances, then couple exerted on the pulley due to friction (Tz ~ 7;) R, where R is the radius of the pulley. Work done by the machine against the couple = couple x angulat ration and rate of doing work = couple x angular velocity. or power of the machine = (T2 ~ Ty) R x 2nn where n is the number of revolutions performed by the shaft or axde of the pate per ound and hence 2zn is the angular velocity of the shaft. Now, by knowing T; and Tz from two Spring balances, R and n, we CA determine the power of the machine, HR, here normal reaction R= Ta. ma pre Vecucnal lore = Miitne elation T/T; = 4 alse band anid the Miecording, mumbar if we know Ts, Ts andl Tr ition m FoR the oa at ight Wie pared on 4 png a >), where) is the B tome , . fe nde of rte x igast force which wilh gevent tag MW down the inclined plane ree which ill prawant vie ¢ inclines slane - ys = plane. If Rs the aers - e pesition of A 7 | ional force will be yt, acting a5 ail ent the body sliding dowuemda : 2). resolving the forces along amo a dicular the plane, we have 5% ~ Wain =P O88 + wR “ wosa=-P sind +R siminating R from €4, (1) 20 (i), we Bet paw SALES - W sin a osb = wsin wotinso For the least value of P, cos (9 - 4) shou wrth, e. (Vaan facies value of P is W sin (a - 2), The direction of P, when it tas eran stan 5 gery jek [: ws@-H=1- 7 La KoA body rests upon an inclined plane and wil ust Side town ir sme } inl won the slope of the plane is OP. Find the accelearicn cf Ine soe town Se Bane hee! when the slope is increased to XP. =i Sol. When the body will just slide down sane, he ncination of he fuse ; by will § at i BW, ie, w = tan 30% When the slope is 60°, acceleration s a=g(sind -ucssy = g (sin OF - tan UP os 11 =98(-#5 aig x5 7 2.3.4 uniform heavy ladder, whose one end 8 z mend ona wall, is in the equilibrum condition. If the youn 4 ih and their coefficients of friction are » and respective, fed the Sol. | with the ground if it is in limiting equal / Vatious| me be a ladder of weight W, inclines 2 ity es tng on the ladder in the concitoe oo . Now resolving vertical and horizontal or pR =R RewR =H andpro | MECMANTES pout As, pe pasin® Ing moment ah raking, } jadder. in) we get Wa cose yt . da & length of th ig WHEE os 8m BR! "CORD cos om? anminating & fron and (i) ye te) = woe aa te) at lt ae or Re ae of Rin eq, (iii), we Bet Weos@ = efi e088 + 8i9 6) Ae substicating the value cose (1 =e’) = 24 8100 ae Laniien. paae HH oro = 100 ( zt } tan @ = Th 1 dings of srping balance Fx. 4. The difference between the readings ances atc hant brake SON. Ip the radius of the pulley i 0.25 m and it makes ‘ot aa find the horse-power of the motor use Poona ye Le or rene ne motor P= (72 ~ Ti) R x 2an. (See Art S-17) Here, 77-1 = 50 N; R = 0-25m dan = 2x22 x49 =2% 22x 7007! © P= 50x 0-25 x 2.x 22% 7I/8 = 3850 J/s. and 5.18. Motion in a Viscous Medium. When a body moves slowly in a viscous medium, the force of friin is proportional to the velocity of the body (¢.g., the movement of abil through honey or castor oil). Thus, this force can be expressed as, F=-Cy ) where C is the coefficient of proportionality. In case, the motion of the body in the medium i al . > ie edium is so fast that thef swirls around (e.g. the movement of an airplane through air), then he frictional drag b a ¥ a velocity i.e., & Decomes more neurly Propostional to the square : F=-C? a or much more higher velocities, even this law begins to fal. oa Flight of an Aeroplane. aac 'ypes of flying machines : sh flying Purpose, oe which are lighter than ait sat Archimedes and it at¢d ait-ships. An air-ship works on th¢ Py "Ses up, because the weight of the 2it di dso is greater than its wei.) NS weight ée,, the upward thrust on it due t9 Harmonic Oscillator “9.1. Periodic Motion. If a particle moves such that it retraces i intervals of time, its motion is said to be peonie Ene eet equal required to complete one round trip (cycle) of motion, is called pont Now, if a particle in Periodic motion moves back and forth over the same path, we call the motion vibratory or oscillatory. To and fro motion of a pendulum, vibrations of a tuning fork, a mass attached to a spring and atoms in a lattice dre the examples of such motion. We shall sec that the displacement of a particle in periodic motion can always be expressed in terms of sines and cosines. The terms harmonic is applied to the expressions containing these functions, hence periodic motion is often called harmonic motion. ¢ 9.2, Simple Harmonic Motion and Harmonic Oscillator. Simple harmonic motion is a particular type of periodic motion and is very common in nature. A particle is said to execute simple harmonic motion when it vibrates periodically in such a way that at ary instant the restoring force acting on it is proportional to its displacement from a fixed point in its path and is always directed towards that point. A system executing simple harmonic motion is called harmonic oscillator. Let us consider a particle of mass m, executing simple harmonic motion. If the displacement of the particle at any instant ¢ bex , then its acceleration will be d@x/dt”. Now according to the definition of S.H.M., we have : restoring force « — displacement d x a) i n—Sa“-x or maT . “ ar de where C is a constant of a at, proportionality. Here, negative sign g oa indicates that the force on the are particle is directed opposite to x_ aes increasing.06 | MECHANICS From eq. (1), we get mfZ4Gr=0 ss. (2a) at If we puto” = C/m , then Er + ora0 (2) it’ Eq, (2) is called the differential equation of motion for a simple harmonic oscillator. Sg . From eq. (1) [F= —Gi, it is clear that the harmonic oscillator is an example of conservative system. At any 14 1 instant, its potential energy Uu=3 (ie. f, Cede) and total energy* is E = mv? +07, a constant. If a graph is and.x, it will be a parabola. Thus, for @ harmonic oscillator, U-x curve or 0 -x-e potential energy curve is a parabola and the Fig. 92 force (F=—dU/dx), acting on it” is 8 7 ‘proportional to the displacement in a direction opposite to it. ‘Now let us solve eq. (2) to find the displacement of the particle at any instant ¢ Multiplying eq. (2b) by 2 dx/dt, we get plotted between U de dey 2a Ht 2a at 2 Intergrating it, we get 2 de 2,2 (4) +P -a . ve Q) where A is the constant of integration, When the displacement is maximum i.e.x = a, velocity dx/dt = 0; 3 . 2 Di OFoWw=A or Amara, 2 (¢) tattade? r da dt This equation gives th . (4), we get NS he velocity of the particle at any time ¢, From de Ve— = wdt. We shall deal in detail in detail the energy consideration of a harmonic oscillator in next articHARMONIC OSCILLATOR | 397 Intergrating further sin” (e/a) at +o or (or + 9) | ; . (w+ aa) where the maximum | ue of the displacement , is called the anplltade of oscil ee and @ is a constant known as initlal phase or phase constant. he term (or + 9) is called the phase of vibration. Eq. (5) is the solution of the equation of harmonic oscillator and provides us the displacement (x) of the particle at any instant t . : In eq. (5), if we put g = +2/2,then x =a cos (w + 6) «+ (6) Thus, ¢ simple harmonic motion may be represented by either @ sine or a cosine function. But the phase constant has different values in two cases. If the time is measured, when particle is at its mean position ie., at x=0,f=0, then from eq. (5), we have g=0. Hence the above equation reduces to x =asinwt 0) Now, let us find the physical significance of the constant @. If the time tine. (5) is increased by 27/@, the function becomes x =asinfot+ 2n/w) + ¢) asin (wt+22+¢) =asin(ot + ¢) That is, the displacement of the particle is the same after a time 22/w. Therefore, 22/o isthe period (7) of the motion, /.€., (. a= £) --® m The number of vibrations per second (1) is called the frequency of the oscillator and is given by .. (9a) - ++ (90) eo w =2an =2n/T The quantity © is called the angular frequency. Alternative method. With the help of the complex numbers we can often solve easily the complicated oscillatory problems, Here, we will make use of complex numbers in solving the differential eqvation harmonic oscillator, Ley Cc ax +@7x =0, where @ = Vv : ae398 | MEC LANICS Let the solution of this equat x Now, substituting the v have t= 0, oF a=+V tie, where, i r=4ie orx= 420°" Hence the general solution of t equation will be ow r= Ae + Are where 4; and 42 are constants, which can be determined from the initial conditions. Now, x A} (cos et + isin wt) + Az (cas wri sin er) = (41 + 42) coset + (41 i — A2/) sin ot Let a sing and (4; —A2i) = a cos. Then, x =asing coset+acas¢sinat or x = asin (wt +g) and ¥ =dv/dt=aw cos (wt + ¢) = @Va* — 9.3. Energy of Harmonic Oscillator. In general, a harmonic oscillator (i) Potential energy, which is d position. (ii) Kinetic energy, which is due to its veloc’ Hence, at any instant the total energy of t Possesses two types of energy : luc to its displacement from the mean oscillator will be the sum. of these two energies. As the system is conservative (e., no dissipative or frictio cs are acting), the total mechanical energy E(=K + U) must be conserved. Below we have also given the proof for the case of the harmonic oscillator. Let the displacement of the harmonic given by of this statement oscillator at any instant ¢ be zr = asin(ot+¢) + (10) Its velocity v= dt/dt =awcos (wt + g=0Ve—-e |, qu) Acceleration d*x/d? = —ea sin (of + 9) = -wx ‘ Force = -mw*x = -Cx [v= C/m} ...(12) If the oscillator is displaced through dr distance, then work done on the oscillator is dW = CrdePARMONIC OSCTLLATOM | 99 Tit is clisplacedd frome = O tow = x then work dh then work done We [Quite s Loe ° ) This work done an the oscillator becomes its potential energy U. Le, v= Je Oh ay Ruretio enengy of the oscillator at the displacement xis 4 py? wx locas? fo? Kofi? = dona? (aay = So?) ay Total energy BevaKk a loateg “2 erg BAK = F004) CaP) = 1 Ca... 18) \ Vita? LinQart a 2 2, nae T? or wna’, Thus, we see that the total mechanical energy is const expect and has the value 1 Ca?, ie, the ti i square _of_the amptinide. Hence, if the system is once osc motion will continue for indefinite period without any decrease in amplitude, provided no damping (frictional) forces are acting on the system, * that if we plot a graph It is clear from the expression U = wie between the potential energy U aud displacement x, we get a parabola having vertex at_r=0 [Fig. 9.3]. The curve for the total energy E (constant) is the horizontal line. ie particle cannot_go beyond the points where this line intersects the potential energy curve because BCTotal Energy) (iba U can never be larger than E.) = Ot These points are fuming points Fig. 93 of the motion and correspond to the maximum displacement. At U energy of the oscillator is wholly potential (/e., E=U= . +t a kinetic energy is zero and so the amplitude of motion is a= as Pee At the equilibrium position, the potential energy is oe “ the 7 ii x a= th maximum energy’ has the maximum value Kanan 2% E wil velcoity Vmax = V2E/m. While at other intermediate points the energy is max hese positions (ie, x=a) the total § Ca?) but theoe Teorey te = MH | MELT ANOS sf ened ) : partly kinetic and partly potential but their sum (tot BY) 18 alway, J 2 a . : Average values of kinetic and potential energies. Average king, enétgy for a period T is given by L(t 1p? 1 in? at Kas = J, Kat = Se 3 csibistiisite ce 1 maoteo? (ot + Ade [Fromeq. any | 7702 ; ; a T= 2n/0 4 | od Sy 510 + 20s ak + 26) ie 2 - ate BE ab maa? = bea? +. (17) o 1 pT 1 (71 Alo, Way =5 S9 Udt =7. S05 CP at = FILE Ca? sin? (wt + pdt = toa? (8) From eq. (17) and (18), we see that the average kinetic energy of a harmonic oscillator is equal to its average potential energy. Ex. 1.A particle of mass 10 gyn lies in a potential field V=502 + 100 ergsign. Deduce the frequency of oscillation. (Agra 1990. Sol. Potential energy of 10 gm U =1073 (502 + 100) ergs. Force =—dU/de = ~1000r, +. 10dx/dr= 100% 2, or £2. \oe=0, de ‘This is the equation of S.H.M. whose frequency is n = VI 5 in =x = 1°6 oscillations/sec, ¥s.2.A particle moves inthe potential energy field U = U; 2. Find he ; = Up - Px + Qe’. Find t expression for the force. Calculate the force constant and time period. At what point does the force vanish ? Is this a point of stable equilibrium > Sol. Force F =-dU/de =+P-29, Evidently, this is a linear restoring force and thi Epis ot ee iB \¢ force constant is 2Q. ae 8 2 | mig Pe ot Sat peak Periodic time = 2Vm72Q: WARMONIC O8CILLATOI “The force vanishes, where dU/de = 0,16, P — 20x = | 40 r= PPO , Cade Now, @°U/d2= +20, be. if O 18 poritive, at . , PD rt potential energy eurve and hence this point isin stable citron mune ot Ex. 3.4 particle is moving with simple harmonic mot i rn wotio in ine dimance of the particle from the equilibrium postion haa tia a mand the corresponding values of velocity are uy and 9, Show that the period is eee T e2af 2A yr Cn ul : (Agra 1991; Kanpur 83) Sol, Let the equilibrium position of the particle be O. If P, il IEP, hi positions of the particle at distances * : eon yy and x2, from O respectively, then the velocity uy of the particle at the ;, position P is given by | =k uy =0Va =m? |.) — i: “The velocity u2 at position ols . Fig. 9.4 uz =0Va? =” (ii) Squaring (i) and (ji) and subjracting, we get 22 =u? (2? x12) my Boag? or o=Vo =e 7 a x7 x1 : ; 2x x? Hence, time period T=Z =2n VW AS : A Ex. 4.A particle of mass 2 gm moves along the X-axis and is attracted towards origin by a force 8 x 10~3x newton. If it is initially at rest at x=10 cm, find (i) the differential equation of motion, (ii) the position of the particle at any time, (iti) the velocity of the particle at any time, (iv) amplitude and frequency of vibration. \ Sol. (i) _ Force mE =a or moe ~8 x10 3x a&x 0 or +4 =! de This is the differential equation of motion. (ii) Let the solution of the equation be x= a sin(2t + ¢) When t=0, x =0-Im, * O-lsasing When 1=0, dxdt =0, ‘0 = 2a cosp [dx/dt = 2a cos(2t + ¢)] a=0 of ¢ =x/2; byt ano, 9 = 4/2 Therefore 0-1 =a sinx/2 ora=0-1mOF FRY vem eS EENOE THEN its, ann ta? Re i pe | 2 Here P20 sec and a=0-16 m an? x a? (otal energy is Biven by SIA x 194 oy ma}0:24 x 1074 y 2 sy wl 2A IOS On ie 9 ER 2en xO 10x 016 Me bx. Vb Af the potential energy of a harmonic oscil in i weil i ion i: joules and the foal energy is 9 joules, when the aplinde is pans Postion ie force constant 1 Af its mass is 2 kg, what is the period ? ema Sol, Potential energy in rest position = 5 joules. : At the maximum displacement (equal to amplitud le), the potential e1 = joules . a ade Gain in potential energy = 9~5 = 4 joules, Ihe gain in energy at the maximum displacement should be equal to } CPrax 2 = 4¢ x ay? = C72 joules. Thus C/2=4 or C=8 joules/m. Period T= 21 Vin/C = 2nV2/8 = 3-14 sec. 9.4, Important Examples of Harmonic Oscillator. (1) Simple Pendulum. A simple pendulum consists of a heavy particle (or ideally point mass), suspended by an inextensible weightless ar. fleable_— left acl CS ‘ . «gos | MECHANH i er bout which the penduty se pom a pone rigid supports aM Pend oy A Wit iction. peed wit ici ig very difficult (0 achieve thes ue Je pendulum, For exp itions of a simp! : co a small heavy metallic spheri al bob, SCS, : Fpeaded by means of a thin and flexible cotton thread, : es a practical simple pendulum, / closely constitul 7 approximate to the ideal one. The thread at its upper end is held between the two halves of a cork, supported rightly in a clamp, ‘This arrangement provides a rigid support and avoids friction, . Let fig. 9.8 represent a simple pendulum with S as the point of suspension, O being the equilibrium position of the bob. If the bob is drawn to one side and then left free, it begins to oscillate about its mean position O. Let 6 be the angular displacement at any time ¢, If m be the mass of the bob, then duc to its weight the.moment of fotce about the point S is . , = —mgl sin 0. If the moment of inertia of the bob (assume it the point mass) about § | is 1, then J=m/? and torque or moment of force =1d°0/de where 2a 22% °0/d¢" is the angular acceleration at the displacement 0. Therefore £0 . 1S = -mgl sind or mie 8 + mgl sind = 0 dt at ao or a + E sind =0 If Bis small, sind = 6 (~ 2, Sr +be=0 « (19) _ This equation represents a simple harmonic motion, whose solution is | given by , 6=Omax sin (wt + $) (20) where w =Vg/1 and ¢ is phase constant. eae [-3-2vF | hayHARMONIC OSCILLATOR | 4a se, the displace ment of the syste alled angular simple harmonic ote |? angular, hence servation of energy, show that the neni pal "® show that the angular speed d0/di of a a 2 v r Fe = |S AE ~ mgt - f fo. gl — cosy} bt) energy of oscillation, Land m are te ma ' i L ngth and oy is ancular displacement from the vertical (Agra 1993, 54,78 \ ential energy of the simple pendulum at any position P reletene y | el ¥ ion O of the bob (Fig. 9.8) is obviou! . sy given U =mgil - I cos 6), % Ob is displaced through a vertical height (I — energy of the pendulum at the postion Paes a. ace Kadi? =F mil aa/dey [ev (=rw)=t d6/dt} total energy E=K+U = jm? (do/dt>? + mel 0 - cos 8) id be a constant, according to the law of conservation of energy. | do" * v2 ee Fr ana : mi simple pendulum of length | and mass m is oscillating with a maximum splacement 80 radians. Show that the tension in the string at the angular ent 0 is ng (3 cos 8 — 2.COs 0). t the bob P of mass i be oscillating about the 1 the angular displacement 8 of the pendulum, mg can be resolved-into two components and mg sin @. The latter component provides ring force for simple harmonic motion and the mponent reduces the tension T of the string. the necessary centripetal force for the rotation pendulum in’a circular path of radius / at the displacement @ is T-mg cos = m/1 @ y is the velocity of the bob at P. ptal energy of the particle =mg ¢ to O) ON (at Po =mgl(1 ~cos 60) =Mg/(1— 0s att mv? (at P). 2g cos 9 — 2ing cos 69 =v wi . From eq. (i) and (ii), we Bet i Vesey. mg cos 8 — 2g 0086 T23mg cos 6 ~ 27g.c0860 ‘ (3.cos 8.290800) = mgcalled the natural frequency we allowed to fall on the open end gf the in the condition that its natura) frequeag tbe incident sound wave. To detea & 2-3 cm length. If the copa ne wavelength i0 which this system wi 25x10 *x2x16 -258 ji eee ax(0-01) ’ 5) LC Cirenit. Like mechanical systems, electrical circuils © property of charge ascillar mee | ancketrical stem oscillations. LC circuit is an excl a i of inductance L is connected in series 4 psc c (Fig. 9.18). Initially, the condenser hee oe see » Now, with the help ofa key the connection © 19 the coll Suge Cut off and the condenser is connected HY Ppese that ai any instant the instantaneous Vill® wp, then Vi2 — w?, say 8, is a real quantity and then from eq. (10), we have x = (ATOPY 4 gy tbe +++ (15) As k>B, both quantities of right hand side decrease exponentially with time and the motion is non-oscillatory. Such a motion is called dead beat or aperiodic and its main application is in dead beat galvanometers. (iil) Critically damped case, If k = @o, then from eq. (10), we have x =(A1 + A2) eae, where C = 4) + A>. In this equation, there is only one constant, hence it does not provide us the solution of differential equation (8) of second order. Now, suppose V2 — @ z =h,, which is a small quantity, Hence from €q. (10), we have ee + Ay hy =e TAL + het )+B(L-At+...)] Neglecting the small terms, Containing h? and higher powers off, we get Se E+ 42) +h (Ay ~ Ady] x eH P+ on 0) where P=) +4) ang O=h (4 ~ 43), or aeDAMPED AND Pe) RED HARMONY OSCTLATOR: is | 478, de oka ° 7 +enk die Othe di ao! a he displacement of th fie © particle is.x = xq and the x =P and ae =O — KP 20 ~ bry oy ‘ X = [0 + (¥ + kro) en! ei an equation represents that initiath ee ly the displ ihe factor [30 4 00" + x0) t] But as time elapers vases ue comes relatively m il apses, the ¢: i vert beet from the Tainan Important and the displa ae continuous | : mum value to zero and (isplacement returne not just occur. uch a motion is called crtcaty donpet bet ‘ ly damped or just 0 t 4 103. Power Dissipation. If a particle oscillates in a medium, then due to the viscosi f : 5 the vis pedi damping forces act on the particle in a Tiedt ceclete is ovement. In this pa work is done by the particle in overcoming the resistance forces. onsequenily, the mechanical energy of the vibrating article continuously decreases so that the amplitude of oscillation becomes less and less. Here, we want to find a relation for power dissipation (4e., rate of dissipation of energy). ‘At any instant ¢, the displacement of a damped harmonic oscillator is given by x =a0 e7™ sin (ot + 4) :. Velocity of the particle &. ap e7# [—ksin (ot + 9) + 0 cos (wt + 9)} Kinetic energy of vibration K = }may? 724 (2 sin? (ot + 0) + w cos? (at + 4) —2ke sin (wt + ¢) €08 (ot + #)} ++ (18) Potential energy U = $C" =}mao?? [v=o] =}mag2e foo? sit’ (ot +o)} ++ 09) The average total energy for a period will be the sum of i at of kinetic and potential energy. If the amplitude of oscillation doe! Pe change much in one cycle of motion, then the factor ¢ pach a constant. Now, we Tet with the : in? "cos? (wt dein (ot +9) e08 (a4 +) whose si? (ar + 9) cos? (ot +) a4, * oe average values for a period are } : ; and 0 respectively.