Ap TER OSCILLATIONS 141 MOTION 1. \\iuat is periowtic motion ? Give some ofits examples RIODIC jotion, Any motion that repeats itself over gain at regular intervals of time is called periodic Jy har nionic motion, The smallest interval of time after whieh the motion is repeated is called its time period. |The time period is denoted by T and its SI unit is and over second I es of periodic motion {) The motion of any planet around the sun in an clliptical orbit is periodic. The period of revolution of Mercury is 87.97 days. (i) The motion of the moon around the earth is, periodic. Its time period is 27.3 days. jotion of Halley’s comet around the sun is: (iti) The m ‘ars on the earth after every periodic. It appe: 76 years, (ie) The motion of the hands of a clock (©) The heart beats of a human being are periodic. The periodic time is about 0.8 second for 3 normal person. is periodic. MOTION 142 oscntatorY OR HARMONIC Give some of its 2. What is oscilla emmples, ory motion ? es back and forth tion If a body move h ion, its motion is said t0 Meatedly about its mean positic a motion if motion. Such atory or vibratory or harmonic 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 () The swinging motion of the pendulum of a wall clock, (ii) The oscillations of a mass suspended from a spring The motion of the piston of an automobile engine, (iv) The vibrations of the string of a guitar. (0) When a freely suspended bar magnet is dis- placed from its equilibrium position along orth- south line and released, it executes oscillatory motion. (i 14,3 PERIODIC MOTION VS. OSCILLATORY MOTION 3, Every oscillatory motion very periodic motion need not be 0s ssarily periodic but latory. Justify Distinction between period mit jotions, Every. oscillatory motion is necessarily riodic because it is repeated at regular intervals of it is bounded about one mean ry periodic motion need not be the earth completes one is not a to time. In adition, position, But eve oscillatory. For example, evolution around the sun in 1 year but it Ad fro motion about any mean position, Hence its motion is periodic but not oscillatory12 ‘AND PERIODIC FUNCTIONS FOURIER ANALYSIS win suitable examples Og scion, constr ction witht peri a Farid Runclion. Ary funet id ment seas terol of aren Ft consider the funtion J property, 44 foeT=f(®) cates that the value of 4 This indats that the tr increased OF ins same when of sceued by fan integral multiple of T fOr all ee ar tanction f satisfying this property OT sic periodic havinga period T. For example, WEA Penctions like sin O and cos Oare periodic W ap of 2x radians, because sin(@ +2n)= sin ® cos (8 +2") = cos 0 If the independent variable @ stands for some dimensional quantity such as time t, then we can construct periodic functions with period Tas follows : ant 2Qat fl)=sin2=! and g,(t)=00s Th We can check the periodicity by replacing t by t + T. 6 us re function f fttTesn tts sin( 2 25] Similarly, g,(t+T)=3, (t) It can be easily seen that functions with peri period T /n, where n=1.2,. iso repeat their values aftr atime lence it is possible to construct two infinit periodic functions such as Nese of . 2nnt fy(t=sin tt n=1,2,3,4,. 2nnt #) = cos 22 8 (1)= 08 n=0,1,2,3,4,.... In the set of cosine functions we constant function g, (1) =1. have included the The constant function 1 is Tand hence does not alter aeatatic for any value of © Periodicity of g_ (1) 2nt F (1)=By +B 08“ +b, cos $81 ant T +b a a, sin et ea sin tre | ost r T *4, a perwes aby +B, £08 OF + by COS 2a noe aysinot +a singe 2% hin, : f=by + Eb, cos not ha eriodiC or F(a b+ + Ee say oe where o=2n/T. wel ‘The coefficients bp, By, Byun, pate Fourier coefficients. These cocttiigae=% r mined uniquely by a mathematical <0 Fourier analysis. Suppose all the Fogy except a, and b, are zero, then 2nt F(t)=a, sin = + by cos 28! ‘This equation is a special periodic simple harmonic motion (S.H.M.) 14.5 PERIODIC, HARMONIC AND NON-HARMONIC FUNCTIONS 5. Distinguish between periodic, ham non-harmonic functions. Give examples of ech Periodic, harmonic and non-harmonic func Any function that repeats itself at regular inter argument is called a periodic function. The folloviag = and cosine functions are periodic with period . 2nt sin =! T Motion Fi f (t)=sin ot Qnt and 8 (t)= cos at = cos 2! T __ Figure 14.1. shows how these functions vay we time f. +1 bo a 5 TR T 3m). -1 @ +1 + so > TOIT -1 ® valor viously these functions vary betwee ow + and minimum value - | passingasc htATIONS "in harmonic functions are necessarily period penn junctions are not harmonic, The Perea ge ili e rindi sevot be represented by single sineor ss functions y cca i sine or cosine shite non-harmonic functions. Fig. 14{ftio" aime periodic functions which repeat tem 2 shows wd T but are not harmonic. selves in a no oT On 3 a Fy 7 Feo] ip, 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(#) jumbo v= aepduy — apraqdury ‘ u : u a+ pO vO = x —— aed ay puns fennm st uonow ayy Moy Uo spuadap anrea coy upsuoo aayisod B sty] ‘UOROUI axp Jo apnyzjdiuw ay Gquenb ay ‘appared ayy Jo y90d2 40 junysu02 som jour paqte> st% pu ajsuied axa jo aseyd psy +yo Ayquenb ayy, +7 queWsUT Aue Ie WHS yard e yo yuowaoejdstp saat8 uonenba sryy (+10) 500 y =x x0 (4+ 10) soo = 10 z ‘) (+ 19) 09 = 22 ° +1 = NOd7 “ANOV pa[sue-yy3is UT tony ‘NO sj queysut Aue qe 4 N uoypoford Jo juawaoeidstp ‘| (°6+) yoda “yu's ut quaurasetdsta 9yT SIE nf . * Sng 14 = goy7 yeup ypns g od aH Pag IO r es ug 28ddns 7 oun Aue vy “89 = VOX? 1 Dang 3PRred aouazayaz aup “= 1 IB “Psue A Pue Y snupes Jo apap # UOTE 550 SSyngENR uo uorsanp 2s}9PoPUE cae BPISUOD "PPL “Bey UE UMOUS "4 aiduns ur quawacetdsid “pm suey Cyr SL “souziafu fo jou paqye> st saajonas q apsuued axp quowareydsig YY Buoye app ayy pue apyied Sujeusue8 20 2psHavd aouauajes pares si q apouued ayy -2jauo v fo sajoump v uodn wo}out seynsut9 uuiofiun fo uoxzaload ayy So pousfep 2q vw yonour suouey afdunts asuap tuoWey ajdunis 9q 0} ples St Q Moge NV Jo UOOUE YL |XX J}9WeIP atp Suoze Q yutod ayy noge o4y puke 0} SaAour NY ‘BPI aup Jo aouazaywmorp aug Suope saafoaar g se ‘SUL “X 010 PURO 04,X “4X 010 ‘001 X WY sarow N £X 1K PUR OL, X 4X 01 {/K01 X WO aUD 249 SuoTe Serout d SV XX Wwe ap uO g Jo voHre/oud a Parte? SIN USL, XX JOIaWeIp ayy 0} g Tod axp wos} Laesp sejnorpuadiad ayp Jo 300} ayy aq N 3°] “@ AIDopeA dejnBue uuojrum pia y srpes Jo apap v Suope Surour g apured e rapisuoo ‘pI “Sig Ur UMOYs Sy ~voHOUT Gejrou> wuOyIUN PUE WH'S weemuieq uone|ay “WeH'S at apyand v fo juowoonydsip ayy 40f uorss -audxa uv aauap 2209p] ‘apuws ayy Jo ssjauvip v Suope Hoyou avprouy wuofiun Jo uoysalosd ayy sv popswSau aq. fu wo4jout suuowiny 2ydtuss Joy) MOUS “OT WH'S NV NOILOW BVINDUID WHOSINN 6°PT ‘apyied 1a ayy Jo UOROU! Jo AES [HME ay INOGE sla} anode n: sed ent payee st aueisuo> ayy %=6 O=FV sypoda = 4 uy 0} Stupuodsowe> arsed ‘Suryougia w fo asond HLL sypoda x0 aseyd jenny (*) . paso atp Jo UOROUE jo UOHDANIP ay) PUE “ee 1 passt “go aso og} pat? $0 0 [+] Che mse aE qe] 0 Ramo yy prs oa ® yt uonesajanze yo anfes uunumxeul au) i sm == uogesajaoe ‘os Y= 1 uo|a000 Hy) WHY SMOYS A TWH'S apaed e yo uoqesaypooe atp sassaudxa uonenbs sei x9 -= (i + 70) 309 F 2 = (%9 + 40) 309 8-= Dd = apyouuerp wo Da 70 WORTH = ped souosaja an jose ayy yo uoRalosd = } quesur sue weg apnied yo UORELa|AOW = aan jo wonei9paove snosueeIST dims st fo woRoW AL ypuadiad DO Pur dd “HC 10 pws ur aygared & Jo WORRIES jrouyve ¥ 40 NOUNET? __ ua eePti-——— 14.10 Displacement, a on = 07 cos 3-8) (E or x(p=beos( 3 a {1:08(-8) = 08 81 or x (p= boos| 5 period 30 5 ‘This represents S.H.M. of amplitude b and an initial phase =~ /2 rad. Example 4. A simple harmonic motion is x=10sin(20#+05) Write down its amplitude, angular frequency, fequehch time period an intial phase, if asplacement is measi el ™ metres and time in seconds. {Himachal 09C] Solution. Given x =10 sin (20 f +0.5) Standard equation for displacement in SHM is x= Asin (ot + @) Comparing the above two equations, we get (@ Amplitude, A=10m, [ Aand x have same units] represented by (ii) Angular frequency, = 20 rad s- @ _20_10 iti va 2-2 10-338 He. (i Freawensy.V=95 35 2x i) Time period, an . (2) Time period P= = 39 10 (Initial phase, , = 0.5 rad. Example 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 T=25, A=Sem, t=1/35 (i) For the particle starting from mean position, Displacement, x= Asin ot = Asin 2% ¢ =5sin 2% 2 _n Example . A body oscillates with stny equation “ng osc x(0) =5eas 2nt-+ x 4) corel auhere ts in sc. and x iN Mees. Caen given bY Displacement at t =0 0 Ti pin see @ Disp me Tn poy, tint (©) Initial velocity (ene Solution. Given 20) =5 coset y ™ We compare with standard equation, x(t) = Acos(at +4) Co (a) Displacement at t =0, and (b) Clearly, @= pe ‘Time period, T=18. s a (0 Velocity, v= a Ssin( 2a ¢ Initial velocity at t =0, m_ Wn v =-10nsin— =~ aes Example 7. A body oscillates with SHM accrngy, ” equation, . Ex A x =(5.0 m) cos [(21 rad s )t+n/ 4), an At =15 5, calculate (a) displacement, (b) spd » th (©) acceleration of the body. wom si Solution. Here @=2n rad s“!, T =2n/o=15 t=15s (@) Displacement, 1.0 cos (2m 1.5 + m/4)=5.0 cos (3x2 .0 cos / 4 =~ 5.0 x 0.707 =-3.535 m. (b) Velocity, dx d “ae gy 80 cos nt + x/4)) =-5.0x 2nsin 2nt + 2/4) 5.0x 2m sin (2nx 15+ 1/4) = ; 2 om =+5.0x2nsin n/4 5.0x2%5 x07 =2222ms1, (©) Acceleration, dod dt dt =~20n? cos nt + x/4) =—4n? [5.0 cos (Qnx 15+ 2/4) =~4%9.87 x (-3.535) = 139.56 m st a [-10n sin 2nt + 1/4)]sc taTiOns ple 8 The equation of simple harmonic my gxomPl? 6 sin 10x t +8 cos 10x t, where y fo tion ty I Bermine the amplitude, jis in om and Period and initial phase Gowwtion. Given y=6in 10% +8 cos 10_ (a) general equation of SHM is yoasin(of + 8)=4 sin @t 60s $4 a cos wt =(a.cos 4) sin wt + (a sin 4) cos «ot sing The 2) comparing equations (1) and (2), we get vo a 8) (4) and Ont or @=10n ‘Time period, squaring and adding (3) and (4), we get a? (cos $+ sin? 6) = 67 + 8? =36+64=100 o a ‘Amplitude, @=10 em Dividing (4) by (3), we get tan ¢=8 =1.3333 6 Initial phase, @= tan (1.3333) =53°8', Exomple 9. A particle executes S.H.M. of amplitude 25 cm and time period 3 s. What is the minimum time required for 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=25 0m, T=35, y=125em 28 28 agg! 125=25 sin 2 ¢ 2n 12.5 _1 or sin2%y = 125. 3 25 Qnt on Bat For 3. 6 4 Time taken by the particle to move between two Points 125 om on either side of mean position is given bY 2te2xd ay 58. 42 Sxample 10. The shortest distance travelled by a particle executing SHM from mean position in? sis equal to(v3 12) times its amplitude, Determine its time period. Solution. Here #=2s, y=(13/2)A T=? Qn y= Asin at = Asin= sin at = Asin = As 1411 CF T=2s, inten gai tinny oe te ln 9 Fiepee Scion The dpa SM gen by Y= Asin (at + 4) - Given T=2s, A=5cm * period ofa simple pendulum is? s fom eulriunposton at mn the stat ofthe motion the pendulum isin imum displacement towards the right of tion, then write the displacement equation sin(xt+@) ), displacement sin (nx 0 +4) Qeani2 Hence displacement equation for the pendulum is ‘cm. Therefore, or =5sin{ rt +)=5 cos x 1? sin(x F\=seosnt the particle to move between two points 12.5 cm on either (evermipte 12. A particle executes S.H.M. of time period 410 seconds. The displacement of particle at any instant is given by :x =10 sin ot (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 08] Solution. Here T= 105, Velocity of the body 2s after it passes through the ‘mean position, v=10(24) 10 = 21 c0s 72° =2 x 3.14% 0.309 = 194 cms. (ii) Acceleration of the body 25 after it passes through the mean position, 2 @) x10 so ‘) T) T 10 sin 72° cia (oy 4x 9.870.951 __ 3.75 ems.te a vemenxo As v=o stein SM te ee a oo example 13 Fay non weayAod or Baa, oe prea afin of te TS e- a y fAPax2 or aug whe reas AD and = 0 YAP ban aoe Hem xicincnantrisin e008 Ly baton As ite to travel ro pa va es Hein a pee OM subtracting om wet en ga om ada Wag, “ To Abe ty Bind bt * Solution. Here x =0 at =0. a Also 2m on THis At bat, x= A/2. Therefore, . A AL Asin@sh) oF 27 Aenean example 16. If the distance y of a pein : ant,== or Straight line measured from a fied origin oy ae Oareconneced bythe relation Wea EN Exo s vvtion is simple harmonic and find iyo Time taken from x=010 = AIS ‘line amo ue Solution. Given 40? =25 ~ y vt or Also velocity in SHM, v= ‘Comparing the above two equations, we fn the given equation represents SHM of amplituis Exomple 14. Ina HCI molecule, we may treat Cl to be of and @=1/2 rad s infinite mass and H alone escillating. If the oscillation of HCI molecule shows frequency 9 10s", deduce the force constant, The Avogadro number =6 x 10° per kg-mole Time-period, Example 17. A particle executing SHM along su Solution. Frequency, v=9« 10's line has a velocity of 4 ms“ when at a distance3 mjue A ‘mean position and 3 sits” when at a distance of mh M Mass ofa H-atom, m= 35 == kg Find the time it takes to travel 2.5 mom th posit _ ity of its oscillation. f eee eee Solution. When y,=3m, 0,=4ms! Vm 4x2 m 4 aw When y,= 4m, 0,=3ms 22 2 -4(2) 1 «(9x10 As praya’-y yes 7) 6x10% mt or 16 = w (A? -9) = 533.4Nm™. = 2 42 eo (A -1) Example 15. A particle is moving with SHM in a straight and Se kee line. When the distance of the particle from the equilibrium Dividing (1) by (2), we get position has values x, and x, the corresponding oalues of 2 ' zeloites are, and uy. Shoo that the time period of oscilation 16 A916 A -250-9% is given by 9 A -16 or 7 A =256 -8 A= V5 =5m Solution, When x=x,, v= From (1), 4=@ 57-3 =ox4 When x=, ven is rad sssc HlATiOns pan the particle is 25 m fom the py wh it its displacement from thera sence x 10 oe resis cm from its equilibrium position at ate pest er ana frictionless surface vosaseasm sins ur: fom et t= tees noted from the extreme posit nthe time Solution. t when the # Here m=1 kg, k=50 Nav!, AAO em =0.10 m : can wile y= Acosat r | 25=5c0s(1x!) on [E= F -707 mast 25 {As the motion starts from: : ence f= 5 ah aU)eAsinet or 2(0)=010sn7471 Eon ¥() =F 0.10% 707 cos 707 + own! —_ ee er aeeclon sow SC hee (8) = 0.707 087.071 ms" a smaximum_velocity 1 and a maximum accele- Exar od waa sus’. Find its amplitude and the period of See ee eas aon a piri tints ny end dry harmony aon: Maxim velo, fering 20 en epi gies the person's weight against time, a soA=40ems? deuce the maximum and minimum rea nae OAM: ported er ings i will sho. Maximum acceleration, Solution. The platf nw? . olution. The platform & 4 that pax = A= 50 6m 8 vibrates between the posi- A ator Acs tions A and B about the | mean position O, as shown “a in Fig. 14.1. = Given A=5.0 cm, i ight m=60 kg v=2 He ne At A and B, the t mae acceleration is maximum Be Period of oscillation, and is directed towards the na Qn _2x3.142«4 ‘mean position. B. a 25.038. Ieis given by = =o0A Example 19. The vertical motion of « huge piston in a Fonax = 0 2A Fig, 16.11 nachine is approximately simple harmonic with a frequency u a a f0.50 s"'.A block of 10 kg is placed on the piston. What is 9.87 x (2) x0.05 =7.9 ms am anplitnd ofthe piston’s SHM fo te Hock [At A,both the weight mg and the restoring force F mM ‘and the piston to remain together ? are directed towards O. Therefore, the ‘weight at Ais Solution. Here v=058", 8=98™ s? ‘maximum and is given by Q) an Wy =(rg + F)=(0g + gy) = 8 fa) ‘The maximum acceleration in SH is given by = 60 (10 +79)=60% 179 =1074N Bax = Az(nvy Acar A 1074 1074 son ag t The beg i remain in contact with the piston 3 0 Je Ogu $8 08 AE ASE ‘AtB mgand Fare opposed to each other so hat the Fence the maximum amplitude of the Piston willbe weight is minimum. Its given PY Ag = 85 0.99 m. W, =(mg ~ F)= 1 = Mlggy) = (8 ~ ina) x gay an” (05) = 60 (10 ~" 9)=(60% 2.1) N=126N Example 20. A block of mass one is fastened spring 218 ano kg & 1. the block is pulled 10 @ 10 with a spring constant 50 Nm.exemple 22. A body of mass 0-1 hs & executing SHM according to the equation 3n) cos( 100 t+ = | metre 3 cos(100 1+ me Find (i) the fequency of oscillation (i initial phase Gi ima eecty io) maxima aceeration and (o) total energy i (100 +28) Solution. Given x =0.5 cos| 1001 +> }metre na) For any SHM, x= Accos(ot +) Comparing the above two equations, we Bet 3x A=05 m, @=100 rads, 4 (Frequency, v= (ay ial phase, = 22 (Gil) gay = 9 A= 100% 05 = 50 ms”. (0) nae =" "A-=(100)? « 0.5 = 5000 ms (v) Total energy 1 1 2 2, = 1 «0.1 x (50) = 125 Je 3 Mam “2 Gay P roblems For Practice 1. Asimple harmonic osilaton i represented Py the Gquation, y= 040sin (4401 + 061) Hovey and tare in m and s respectively. What 2% Bers Sues of () amplitude (i) angular frequency thy frequency of oscllatons (@o) time period of bscilations and (o) inital phase ? [Ans. (@) 0.40 m (if) 40 rad: 7 (ii) 70 Hz (io) 0.0143 s (x) 061 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. 1/6s) 3, A particle executes SHM represented by the equation: 10y~ 0.1sin 50 x, 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 = 40s" (t / 2)sin (1000). Find the independent constituent SHM's. UIT 93] ; [Ans. sin (1001) sin (1000) sin (999¢)] 5. A partic Ape ee a oa completes 1200 oscilla- pie ee a through the mean position with a velocity of 31.4 ms“. Determine the maximum displacement of the parce Tm the wean postion. Also obtain. displacemen, equation of the particle if 6 displacement be 2er at the instant t = 0 fans. 4=0025.m, «nese te tb cna: dune inept re 2 7 eer 1 \um velocity. eee an " ee (Ans. 0.1571 ms") ster its motion begins, will 2 partie oscillating me eeding to the equation ¥ = 750 05 5 oem. the mean POST to masini ‘displacement ? te oe as cle executes SHIM on a tt path The 9. Apt sean 2m. When he amp particle rom the MEAN POSEAN islon ment Oigritade of ats acceleration © equal to is the maBEnd the me period, maximum veac eceleration of St (ans. 3.635, 3464ems™, Gams to. The velocity of a particle describing SHM 6 seeps ata distance of 8 cm from mean poston leamgem sat a distance of 12cm from men position, Calculate the amplitude of ‘the motion (Ans. 1306 11 A particle is executing SHM. If and st ae Be Speeds ofthe particle at distances, and ftom, trilibrium position, show that the frequency « oscillation, fand maximum a 12 3\. =X, ) {Central Schools it 12. Ifa particle executes SHM of time period 4s amplitude 2cm, find its maximum velocity and at half its full displacement. Also find ‘acceleration at the turning points and when (Ans, 3:dms displacement is 0.75 cm. 272em 5", 13. Show that if a particle is moving in SHM, itsve™ ity at a distance V3 / 2of its amplitude from the tral position is half its velocity in central post [Chandigarh 03 ; Central Seheols 14. A particle executes SHM of period 12> seconds after it passes through the ome oscillation, the velocity is found to be 31° Find the amplitude and the Tength of the PP ‘A block lying on a horizontal table "" period 1 second, horizontally, What ©(0) Time period, T 1 oF ant = jitude for which the block does not slide > Falcon of tition between block and gyri . pate id (Ans, 98 on) sizontal platform moves up and Atrora the fl verwal mover eee 10cm, What is the shortest period permissible resting on the platform are to remain in contact with it throughout the motion > Tate pe gocm s. (Ans, 0.449 5) In a gasoline engine, the motion of the piston is simple harmonic. The piston has a mass of? ky end stroke (twice the amplitude) of 10 em. Find maximnum acceleration and the maximum unbalanced force on the piston, if it is making 50 complete vibrations ‘each minute (Ans 1.371ms?, 2742 Ny 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 wi machine ? The amplitude of vibration of the platform is 5 cm. Take g = 10 ms~. (Ans. Max. reading = 89.5 kg f Min. reading = 10.5 kg f) HINTS Comparing y= 0.40sin (401 + 0.61) with y= Asin (ot +4), we get (Amplitude, A = 0.40 m. (s) Angular frequency, w = $40 rads? @ _ 40x7 (ii) Frequency, ») Initial phase, 6, = 0.61 rad. 2 Here T=28 y= A/2 t= As y=Asinut = Asin 1 T $e Asin 1 Asin xt =sin= or ai=* 278% 6 ba1/68 Hoon? (6/2) sn (10001) 1+ con +) sin (10004) 2'sin (1000 ¢ ) ~ 2.sin (1000 ) cos t 2 (10M 4) = fxn (O00 ¢ = #) + sim 1000" = 17 2am Aces Bean( A+ B- sn(A~ B)) sn (1001) + in (9991) ews? fe 1+ cos 20 2 (1000 +) Thus mot ion vis composed of three SHMs which are ‘sin (1000 ¢) sb oe (1000). sm (200tT Here Sime = 01 so we can write ¥= Asin of = A sin (2x vt) = 0.025 sin (40x). Here @ = 12cm y=3on R 1 Yo "V3 7284s Tain 05 xt On comparing with the standard equation, Y= asin at,we get: 4=7,0=05% Let tbe the time taken by the particle in moving from mean position to maximum displacement. Then 7=7sin O51 or sinOS xt = 1=sin osnt=% or t=—t 2 503 s Here A =2cm. When displacement y = 1m. ‘magnitude of velocity = magnitude of acceleration or or or V3 x2 = 1.732 x2 = 3.464 ams) fant? An3xdegemst, As v=eya’ In first case: 16= 0A ay In second case Dividing (1) by (2)Here T=4s, A=2cm, wo 2% 22% 3-14 157 rads 4 Uyay =@ A= 157 %2= 3.14 em 51 At A/2= 10m, =2720m ono (aay? «157 = [At the curing point, acceleration is maximum Opax = 0? A =(157)? x2 = 4 at ynazsem, : a= 0? y=(1.57) «0.75 =1.85 cm $7. Here y=V3/2A v=o fA?-9 =0 (A348 -joa 1. Let y=Asin ot dy 2n 2n va. = 2% Acos Et Then v= =o Acos ot == $ 2x3142 , an Acos 2 x2 2 or A.=12.cm and length of path = 2A = 24cm. 3.142 = 15 gay =O? ARH g ug gt? aahg abst 0.4 x9.8 x (1)? <= 0.098 m =9.8 cm. 4x10 : 2 6. Take ta, =0? A=( 2) Ang. 16. Tak (2) 8 17. Length of stroke = 2 A= 10cm. 18. Here m=50kg, v=2Hz, «9.87 x2? «0.05 in? y? A Max. force on the man = m(§ + yay) = 50(10 + 7.9) = 8950 N = 89.5 kg f. Min. force on the man = M(8 ~ dyyqx) = 50(10~ 7.9) = 105.0 N =105 kg £. 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 HM. In which op positions of the oscillator, is the energy wholly kinetic or wholly potential ? Total energy in S.H.M. The energy of a harmonic oscillator is partly kinetic and partly potential. When a re body is displaced from its equilibrium position by doing work upon it, it acquires poten the body is released, it begina ty mal . velocity, thus acquiring kinetic enerpy"® (i) Kinetic energy. At any instar, sy, of a particle executing S.H.M. is given ca X= Acos(ot +4) Velocity, 2-7 =- © Asin(on sg, dt Hence kinetic energy of the displacement x is given by 1 io? K => mv 2 Pattie ae ne? 42 mgm A sink toc But A? sin? (ot + $)= A? [1 ~ cos? (op , = A&A 008" (ot + 4) = 92 fl a k=} mo? A sin? (ot +4) 1 1 yor or Ka 5 mat (A —2)=1 (at (i Potential energy. When the displacemen particle from its equilibrium position is ,theresnr, force acting on it is % If we displace the particle further through a sy distance dx, then work done against the restoring ie is given by dW =-Fdx=+kxde ‘The total work done in moving the particle fon ‘mean position (x =0) to displacement x is giveny w-=jaw- f rearai[2] 0 z 0 This work done against the restoring force is se! as the potential energy of the particle. Hence potetil energy of a particle at displacement x is given by 5 mat A x(a (iii) Total energy. At any displacement x. the energy of a harmonic oscillatory is given by 1 lea a 2 Usleeal = 7 5 marx E=K+U= k(x hh 2 sia E Sk Aad mo? a a2 ml A Thus the total mechanical energy Of ° illator is independent of time or di? nce in the absence of any frictional force Of a harmonic oscillator is conserved.the mean position, the energy is all kinetic. ce at the two extreme positions, the ene ray is Energies EE, and Eare mete, fore Son ‘gy U and total energy E with displace- — Se ee saa phs for K and U are parabolic while omple 23. A tod straight line parallel to the displacement the energy is all Kinetic and for © 14.12 showss the variations of kinetic energy A=1Wem=010m y Kinetic energy £,= 1k(a—y) =} 50 [0108 - 0057) 01873 | —— Displacement — Potential enegy E,=}hy Total energy 14.12 K Vand Zas functions of displacement x for a harmonic oscillator. Figure 14.13 shows the variations of energies KU EB, +E, =018% and F ofa harmonic oscillator with time ¢. Clearly, twicein Example 24. A body ex each cycle, both kinetic and potential energies assume its massbe0.1k their peak values. Both of these energies are periodic its mean pusition be 4ms functions of time, the time period of each being T/2- (ji) potential energy and (H) total en eA Solution. Here m=0.1 kg. 25 J. utes SHM of time period 8 s If ity 1 second after it passes through find its (i) kinetic: energy But © Aces ot Ey As . “ 4 $12 K Wand £ as functions of time ¢ or for a harmonic oscillator.Hie Pit Example 27. Ata time when the displ amplitude, what fraction of the total ener i, 6 J. what faction is potential in S.H.M, 2" 8 Why! Solution. Total energy beh mo? . Displacement => amplitude op y 1 z Total energy of SHM, E= i meh Kinetic energy, . 1 Kinetic energy of SHM, E, =! 16-08 =08 J. ficenes m5 MOP (AR force constant 800 Nii has art ‘done in increasing the ik Example 25. A spring of extension of 5 cm. What is the work ‘extension from 5 om to 15 om ? [AIEEE 02] Solution. Here k =800 Nm, x, =5 cm =0.05 m, xy =15 cm =0.15 m a ae Potential energy of HM, 2k G3 By pay =} maa( AY = 4x 800 (0.157 - 0.0571 p> a 5 =8). Example 26. A particle of mass 10 g is describing SHM along a straight line with a period of 2 s and amplitude of Example 28. A particle is executing SHM of anlina 10 cir: What is the kinetic energy when itis (j)2 om(ii) 5 om Ap what displacement from the mean postion, ste oe, {from its equilibrium position ? How do you account for the half kinetic and half potential? ; difference between its tv0 values ? 5 Solution. As Ey =Ep Solution. Velocity at displacement y is ve0fa-¥ Given A=10.cm, T=2 or Angular frequency, or () When y=2 cm, Thus the energy will be half kinetic and =n (024 =n 96 cms potential at displacement + con either side o mean position. Example 29. A particle executes simple harman = 490 song of amplitude A. (i) At what distance from the meat, aween is its kinetic energy equal to its potential eners! \ what points is its speed half the maximum spee! =n (100-25 = 25 ang Solution. The potential energy and kine! sc ke 1 2 of a particle at a displacement y are given" : End hy? - 2 Fi = 7% 10x x! x75 =375 nerg. 1 The KE. decreases when the “ mate Particle moves from ” to the increase in the Where A is the amplitude and _ kis the force constant. y=2cm to y=5 am. This is due potential energy of the particle.= 0,71 times the amplitude on either side of mean position. (i Here, => Pax ingeneral, Kinetic energy 2 ; x Maximum kinetic energy _ 1 + a etx an 81m (ns. 028,25) HINTS From equation (1), 1. At the lowest or the mean postion, energy of the netnatey) tb a Act Pate ot athe 1 net At (Emu = Ame? A (Emax = 3 kat [Put y=0] 2 Putting these values in equation (2), we get Lycae—ypyatx tea Ly (at-yyatx Ska’ gh = 49 2 or 4y? =342 or y 28 ans osea =0.86 times the amplitude on either side of mean position. Pra blems For Practice 1. Abob of simple pendulum of mass 1 gis oscillating with a frequency 5 vibrations per second and its amplitude is 3 cm. Find the kinetic enerBy of the bob in the lowest position. (Ans. 4441.5 8) 2. A body weighing 10 g has a velocity of 6cms 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 enerey: (Ans. 180 erg, 540 erg, 720 er8) A particle executes SHM of period 8 seconds. After What time of its passing through Cie ee will the energy be half kinetic and Veheedigath al (Ans.1) 19 The total energy of a particle executing Petod 2s secnds i 1at =] ese ment ofthe particle at x/ 4 1s 08/2 m Cate amplitude of motion and mass ofthe pac (Ans. 0.16 m ; 0.08 ky horizontally with simple harmonic motion with a Freauency of 1/ x Hz and total energy of 10. the qranimum speed ofthe particle sO ms", what is e force constant of the spring ? What will be the ‘maximum potential energy of the spring during the motion ? (Ans, F=500Nm“!, Upge = 10) ‘The length of a weightless spring increases by 2 cm. 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 ence saad mt = 29.87 «1x5? x3? = 4441.5 erg, vee 106. T288 seh ae PTE. i When ¢=1s, v=6cms” As v= Aocs ot Potential energy = Total energy ~ Kinetic energy = 720-180 = 540 erg As PE.-KE. 1 lye sky— Bp horizontal table. Its one end i, an ae 1e other end Lachey scl = Asin support and the otf toa body oO Now yaasin of = 4807 body Is pulled towards right three and released, it stars oscillating pe ah spe its equilibrium position under thy"! ing) mg oF restoring force of elasticity, Aton a Pe—ky the | where kis the force constant (restoring compression or extension) ofthe spring Th" jy 1 indicates that the force is directed opposite thr Equilibrium Feo Po bDOUOTUTTTOT™ 3 " act y k¢ e stretched! n or 00812 = Ax L woossonnnconTi 08 V2 x /2 = 0.16 m. et ie , 1 energy = 4 ma? A? = Gompremed Fenks i" Total energy = 5 LTOOUTTTID- i 1,,(28)" 2 ext x10 24 m(22) (0:16) a ranger? . Determine the value of v, / vy. [Chandigarh 03] (Ans. ¥2) 5. The periodic time of a mass suspended by a spring HINTS (force constant k) is T. ring is in ( ) is T. If the spring is cut in three yy pF _ 10 _ sg ent, La k, Fig. 14.26 equal pieces, what will be the force constant of each y 01 part ? If the same mass be suspended from one piece, what will be the periodic time ? Gy eee oe mn. 1 4m. (Ans. 3k, T/-¥3) a” 6. The time period of a body suspended by a spri mae (eo, | be T. What will be the new perio, if the spring © Par arg 2s 00 cut into two equal parts and when (i) the body is suspended from one part (ji the body is suspended 7) AS T= 2m £ from both the parts connected in parallel, i arn [Ans TIVE ay Ty 7 Aes ense, wo identical springs have the same for me = of 147 Nm“. What elongation will be re an if ltr “r each spring in each case shown in Fig, 14.24? In second case, Take g = 9.8 ms, TAns. (a) 1/6 m (6) 1/3 m, 1/3 m gle ED yeah (c) 1/3 m} Nie ta force, F applied 09. displacements 2, and #, int Fe x a he etinto three equal pts the Free mes Sppended from one such pee wi [mT pune (is, Tee roan | Vk vale Hb 4) id ie Ke hy e syaten, K parts, then the « spring, is cut into two equ int of each part becomes 2h i Lorain i silly dt 1 kik ispended from one pat, its Yvan Vcd 4 14.16 SIMPLE PENDULUM 2 k4 2k = 4k 4 of oscillation bes T rete (ee. Nae can neithe po weightless string, 1 simple pendulum is obtained by long and fine cotton 1 In Fig, 14.240), In practi suspending a sma thread from a 1 position, the bob ¢ below the point of suspe displaced on either side oscillate about the mean position = 147 Nm 2k = 2.147 = 204Nm In the equi nin the spring, BOB DT K 294 6 14.24(h), the effect xk MP ay! K im kek 22 mple pendula jon, IC the bob is slightly wed, it begins to ant, spring con at any instant during Suppose its displace») Force acting on the bob of a pendulum. Bg » has two rectangular components along the thread is thread and (ii) the The force ygential 2. Frequency, v = =~ Units Used Length | of the pendulum is acceleration due to gravi Example 43. What is the length of a sin which ticks Solution. The simple pendulum whi, isa second pendulum whose time perio Time period, T= 2n seconds ? {NCERT ; Del chi, T=2s, g=98ms? the mean ee As Toan ll or Tha’! canon (0-6 -—) 1g _OFH98 yoy, an 4x987—0 0° ™ aay of! Fe } Example 44. A pendulum clock shows accra tin, 6 120 length increases by 0.1%, deduce the error é Where 0 is in radians. Clearly, 0s simple harmonic because the restoring force F is not proportional to the angular displacement 8. However, if is so small that its higher powers can be neglected, then Solution. Correct number of seconds pet da, v=24 x 60 x 60 =86400 Let error introduced per day =x seconds Then incorrect number of seconds per day, F=-mgo If is the length of the simple pendulum, then ee ee If lis the original length of the pendulum, te ant new length will be P'=1+0.1% of I= = (1+ 000)! F some *t00 (+0000) 1 or naz 28 » Now frequency, v= |8 ie, ve 1 anVI al 1 ~g6400 (+ Ol Thus, the acceleration of the bob is proporti portional t its displacement x and is directed opposite to it. Hence 1+ pean = (+0001 for small oscillations, the motion of the bob is simple harmonic. Its time period is = 1-2 0,001 =1 0.0005 rt a fl 2 ee or Teae fh oe rE or *_ __0,0005 Obvi viously, the time period of a simple pendulum OF x = ~0,0005 x 86400 = - 4325 depends on its length J and acceleration due te but is independent of the mass m ‘of the bob. Bravily g The negative sign shows that the slow and it will lose 43.2 seconds per a 2is the shpercentage incneg 1 Avo Ji,s0 von of the vena is simple dm increasing the length by 43 rare wh fr won 1 1a 2+ oy Ra, ‘ 6. On the moon gy = 8 =F ns 2 meer an | Moan |® As Ta2n | Tatey “PT Me/R Vs Ven Tg, 2 x98 yoblems For Practice 10 GP pp ag Om a m to perform ‘The time taken by a simple pendulum (0 p 100 vibrations is 8 minutes 9 secon: ids in Pune. fcuiaaisie 14.17 OTHER EXAMPLES OF SH. Se ei goripinBonty and 20, One eof 4 Usinhe coninny ratio of accelera (Ans. 1.0455) connected to a suction pump and the other ie Pune . 2 here. A small 8 ength ofa pendulum is decreased by 2% find nected fo the atmosp Pressure Arte engstiole pers maintained between the two columns Show the suction pump is removed, the liguid in the gain or loss in time per day. (Ans. Gain of 864 5) 5 Ifthe length of a second!’s pendulum is increased by executes SHM. i 1%, how many seconds will it lose or gain in a day ? Oscillations of a liquid column in gy, (Ans: Loss of 4328) Initially, suppose the U-tube of cross-section Aq. is iner H f density p upto height h. The ‘ length of a simple pendulum is increased by liquid o! en mag * PR wert al . liquid in the U-tube is iu Volume x density = Ax 2inxp 45%, what is the percentage increase in its time period ? (Ans. 22.5%) 5. What will be the time period of second’s pendulum ifits length is doubled ? (Ans. 2.828 s) * 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. 165 em) HINTS Let g, and g, be the values of acceleration due to gravity in Bombay and Pune and 7; and T, be the Values of the time-periods at the respective places, Fig: 14.29 Oscillations ofa liquid column in a U-tube. ___lfthe liquid in one arm is depressed by distance ‘ses by the same amount in the other arm. Ie itself, the liquid begins to oscillate under the restont force, F= Weight of liquid column of height 2) FS-Ax2yxpxg=-2 Apgy je, Fay r Thos the force on the liquid is propor i isplacement and acts in its opposite direction He ‘Quid in the U-tube executes SHM with k=2ApgF er = serio of oscillation is 1431 ime-Pe The period of oxcilation ofthe body seit y Te2myy Ry Tan [Morn she length of the liquid column, then yi th aa 22. A cylindrical piece of cork of b and T=2n jt height h floats in a liquid Se fland \2s iquid of density py. The cork is depressed slightly and then pele then released. Show thatthe con . 21 Utigit hole bored in it through its centre, show i propped into the hole will execute SHM and rime period. Delhi Fe daoppe ina tun n damping det ety ofthe tidy wea Onereter of the earth. AS shown in Fig, 14.3, __OSillations of a floating cylinder. In equilibrium, ac aes weight of the cork is bal: hm e ior earth to be a sphere of radius R and centre O. creme deme ee oe ig’ Where p is the density of cork. (lgnore eu along the donee ad Ath Let g be the value of acceleration due to gravity Sree surface of the earth, ere Position ore 7 m +P) bh ° Fig. 14.31 Oscillations of a floating cylinder. Let the cork be slightly depressed through distance y from the equilibrium position and left to itself. It begins to oscillate under the restoring force, F =Net upward force = Weight of liquid column of height y fig 14.30 A body dropped in a tunnel along the diameter of the earth. or F Suppose a body of mass m is dropped into the Ayp,s=-Ap gy ie, Fry tunnel and it is at point P ie,, at a depth d below the Negative sign shows that F and y are in opposite surface of the earth at any instant. If g’ is acceleration directions. Hence the cork executes SHM with force due to gravity at P, then constant a a R-d gos(1-q)> ( R ) Also, mass of cork = Ap h If y is distance of the body from the centre of the :. Period of oscillation of the cork is earth (displacement from mean positon, then rene are fa on PE. | nae Fr8R Ve Ap, \Pi8 23, An air chamber of volume V has a neck of area of ™. cross-section A into which a ball of mass m can move Fe-mg=-"Sy ie, F*¥ without friction. Show that when the ball is pressed down Ne a ‘ through some distance and released, the ball executes ons *Bative sign shows that the force F acts in the cris’ Optain the formula for the time period of this Poste direction of displacement i, it acts towards cri44. qssuming pressure-volume variations ofthe air writ Postion O. Thus the body will execute SHM 15. isothermal and (i adiabatic. INCERT] ™ ce constant, Force acting on the body at point P is Oscillations of a ball in the neck of an air chamber ame Figure 1432 shows an air chamber of volumesk ot area " i a it be he Bs JI he wick i | ya ttle a - el _ the anal dows about f : rin pesto. : Ir the ball be depress inthe nis AV = AU av AY Volume strain= 4 ~ ostatic 1 to the ball, then hydrosta! 1 pressure P's applied tress = P 1s of elasticity of ait, modu ‘a Samstag gy ‘ AV/V Ay/V ; tay ,.F&y Restoring force, F= PA=— 2A Ae = actin ts opposte dee rn Fis proportional to yand act ton ene the bl ecetes SHIM with force constant i; ve v Period of oscillation of the bal is Pade ney (i) I the P-V variations are isothermal, then E = P, [mv Vat (ii) the P-V variations are adiabatic, then E=y P T-2n 24, Show that the angular oscillations of a balance wheel of a watch are simple harmonic. Hence derive an expression for its period of oscillation Oscillations of the balance-wheel of a watch, In a watch, a balance-wheel controls the movement of its hands. An axle passing through its centre is held between diamond points. A hair-spring con- trols its oscillations, two Fig. 14.33 yr placemer nL 4, 3 ag tongs sown! see tee at enim pn : ge + Cabot tude OF tsomple 93. ws sai 2 Fe a \ yr its mitial wre INCERTL at san 2d WES tae BE gs 0.200) , tay tere var aration. p AES so the: aa gostant, POA ES ig small Dam ing constant OSS 8 sree samping i ne period Tis gi v2K Lass. 2 i =2n ye 28 yoy Nm oe i Je to drop to half of plitad 7, for the amy (0) The time, T its initial eis given by vale is Be Ae Typ ‘Te time, fy for the mechanical energy to drop toa inl vale given y a E(byg)= EQ) CO" E (typ)! E(0)= exp (bo /™) 1/2 = exp (btyy/m) In(i/2)= or = (typ) 14.19 FORCED AND RESONANT OSCILLATIONS : 26, Distinguish between forced and resonarit oscillations. Give an experimental illustration in support 9f your answer. Give examples, jared oscillations. When a body oscilates under the influence of an external perc ‘odie force, not with its own natural frequency but with the frequency of the external Periodic force, is osilations are suid 1c be forced oscil Intions: The extemal agent which exerts the periodic force is called the driver and tho oscillating system, tnder consideration is called the Ariven body, Examples.) When the stem o ing tuni This is because the Particles of table ate fog vibrate with the frequency of the tuning fork,“ {) When the fee end of ah Pendulum is held in hand and th of the drivin, oscillator its l loud sound Of the tuning fork. oy (i) A glass tumbler ot a piece © ibrations whe" * Shelf is set into resonant vibration: covers > v9, But when Plt, Int fiction lator freq ions ig vey, driver Fig. 14.38 Amplitude a of a forced ostilatorasafus: Examples. (i) An aircraft passing wes ters its window panes, if the natur uni the window matches the frequency of thes Sent by the aircraft's engine. e tuber (i) The air-column in a resonance Sung or played. tet when its frequency matcheswr | spd four Pe 3 scot the pendulum A into oscillation It LATIONS mental illustration. As shown in Fig, 14.39, P ndulums A, B, Cand D from an el lastic Pa ree oscillations. The energy from cs i ransferred to other Pendutas ‘thresh ose tring, Initially, the motions of B Cand Dare pect put soon all. these pendulums star cz with the frequency of A. The oscillations of silat are forced oscillations. But pendulums B BC aMgve small amplitudes. This is because the ant gncy of Bs much larger than that of 4 (due to eer length) and the frequency of Dis much smaller ropa of A (due to larger length). The pendulum C has same length as the pendulum A (and hence we frequency) oscillates with largest amplitude. lations of C are resonant oscillations. wing “quam which the samé Hence the oscil Elastic string. P Q D fig, 1439 Mlustrating free, forced and resonant oscillations. 27. Briefly explain the principle underlying the tuning of a radio receiver. Principle of tuning of a radio receiver. Tunit the radio receiver is based on the principl 14.37 14.20 courted OSCILLATIONS, 28. What are coupled oscillations ? Give Coupled oscillations. A syst vs exchange of energy between them is eae The oscillations of such a system are tiled a coupled escann called coupled osciat Examples. (i) Two masses att Es ses attached to each ot spring provides the coupling between the driver cna the driven system (Fig, 14.40(a)} eran (i) Two simple pendul Trig tea Pl Pendelams coupled by «spring (iii) Two LC-cireuits i) Two | its placed close to each other. The circuits are linked by each other through the magnetic lines of force (Fig. 14.40(c)). k Fig. 14.40 Coupled oscillators. When two identical oscillators are coupled togethe ing of the general motion of such a system is complex. It Je of periodic but not simple harmonic. It can be viewed « the superposition of two independent simple harmon resonance. Waves from all stations are present around the antenna, When we tune our radio to a particular motions, called normal modes having angul station, we produce a frequency of the radio circuit frequencies «, and ®, The constituent oscillate which matches. with the frequency of that station, execute fast oscillations of average angular frequen When this condition of resonance is achieved, the radio yy =(0 + @,)/2. The amplitude of either oscilla receives and responds selectively to the incoming varies with an angular frequency (o-o). T ‘waves from that station and thus gets tuned to amplitudes is known station, @,)is called beat freque that phenomenon of variation of beats and the frequency (0, CONCEPTUAL PROBLEMS AESSAN eels Solution. Yes; when a ball is dropped from a hei ‘elastic surface, the motion is oscillatory by mg = constar Problem 1. Can a motion be periodic and not oscillatory 2 a perfectly simple harmonic as restoring force F = circular motion iS | 4'F «c— x, which is an essential cond Every simple harmonic motion is pe dic motion need mot be s p tor St »». Yes. For example, uniform pdic but not oscillatory. Can a motion be oscillatory Perc ne Problem 3 but not motion, but every perio give an. -—-aate matian. Do vou agree ? Give one exartCOMPETITION SECTION Oscillations periodic motion. A motion which repeats itself cover and over again after a regular interval of time is called a periodic motion. ory motion. A motion in which a body moves back and forth repeatedly about a fixed point (called mean position) is called oscillatory or vibratory motion. . Poviodic function. Any function that repeats its value at regular intervals of its argument is called a periodic function. The following sine and cosine functions are periodic with period T. fyesin2® and (1)= cox 2 The periodic functions which can be represented by a sine or cosine curve are called harmonic functions. Al harmonic functions are necessarily periodic but all periodic functions are not harmonic. The periodic functions which cannot be repre- sented by single sine or cosine function are called non-harmonic functions. | Fourer theorem. Two infinite sets of periodic functions with period T are f,(0)= sin 2, 1212.34 2nnt g (t= cos et, 1 =0,1,2,3)-- 81 (t) = cos ” Fourier theorem states that any periodic function F(t) with period T can be expressed as the unique combination of sine and cosine functions f, (#and g,, (1) with suitable coefficients. Mathematically, F(t)=b, + 2b, cosnot + Ea, sin not where w=2n/T. The coefficients by, by, bo» 4d, dy,.... are called Fourier coefficients. The special case of Fourier theorem in which only 4 and b, are non-zero represents simple harmonic ‘motion (S.H.M.). Fa 2nt 2nt sin 27 + by con =. Simple harmonic motion. A particle is said to execute simple harmonic motion if it moves to and fro about a mean position under the action of a restoring force which is directly proportional to its displacement from the mean position and is always directed towards the ‘mean position. If the displacement of the oscillating particle from the mean position is small, then Restoring force « Displacement or Fax or Fo-kx where k is a positive constant called force constant ot spring factor and is defined as the restoring force produced per unit displacement. The negative sign shows that the restoring force always acts in the opposite direction of dis- placement x. The above equation defines SHM. . Oscillation or cycle. One complete back and forth motion of a particle is called cycle or vibration or oscillation. Displacement. Itis the distance of the oscillating particle from the mean position at any instant. It is denoted by x. plitude (4). The maximum displacement of the oscillating particle on either side of its mean position is called its amplitude. Thus Kwan = Asn Hie oilation se noun maton i orveca of tee pend, FNS Y= y secneostred in hertz [ote mt pent 1S 6 the quantity abta " suplying, frequency why a factor of 27.1 denoted by 0 . 28 st unit of w= rad s thus weave” 2 The phase of iating particle at any instant gives the state of the particle as regard its position and the direction of motion at that instant, tis denoted by ¢ 1 The phase of a vibrating ing, to time £=0 is called phase or epoch. It is denoted by 4 © The phase difference between two vibrating particles tells the lack of harmony in the vibrating states of the two particles at any 14. instant 1 between SHM and unitorm circular Simple harmonic motion is the projection of uniform circular motion upon a diameter of a circle, This circle is called the reference circle and the particle which revolves along itis called reference particle or generating particle 5. 16. Displacement in SHM. In a simple harmonic motion, the displacement of a particle from its equilibrium position at any instant tis given by ¥(1)= Acos (ol + 4) Here A is amplitude of the displacement, the quantity (ot + 4,)is the phase of the motion and 4) 8 the initial phase. When the time is measured from the mean Position, x(1)= Asin ot When the time is measured from the extreme Position, X(1)= A.cos wt Hoe angel 8 + ang pend Foti ne are given by toile, 1 anVe ; og (Diplacement "Acceleration ‘ Inertia factor or Tring factor Vi jo SHIM. Ht is the rate of eb " scoment of the particle at any instar, ay 4 FA cos(ot + %)1 ay dt WA sin (ot +4) = Oy The maximum value of velocity is c amplitude v,, of the motion. 2B Ay ‘Thus v,, =@A At the mean position, particle velocity =», Al the extreme position, particle velocity =0, 18. Acceleration in SHM. It is the rate of change of velocity of the particle at any instant. Itis given by fo 24 wAsin(ot+ 4) dt dt A .c0s (wt + $y) =— 0x ie, axx The maximum value of acceleration of particle is called acceleration amplitude a,,. Thus, | 4, = OFA At the mean position, particle acceleration =0 At the extreme position, particle acceleration 4, = 0A, 1. Phase relationship between displace velocity and acceleration. In SHM, the particle velocity is ahead of displacement by x/2 rad while acceler ion is ahead of displacement by rad, 20. Energy of SHIM. Ifa Particle of mass m executes SHM, then at a displacement x from mean Position, the particle possesses potential and kinetic energy,lo. ren 2B. ULATIONS (Competition Section) Atany displacement x, potential energy, ode! | Kinetic energy, K= Emu? 4? 2y = kay ‘Total energy, 1 E=U+ K=3 me? A? 2292 my? yt If there is no friction, the tot energy, E=K+U, of the s remains constant even though k tal_ mechanical system alwe = and U change spring, When a doa male es spring eed pull downwards, it exceues spre g and Extension in the spring on attaching el {is its force constant, then time period of SHM. tnecited by the spring i Motion of a massless loaded mass m is attaches 11 into parts, If we divide the spring of spring constant k into 1 equal parts, the spring constant of each part becomes 1k. Hence the time period when the same mass mis suspended from each part is m \nk Te2n nected in series. If two springs of spring constants k, and k, are connected in series, then the spring constant k of the combination is given by If two springs of spring constants k, and k; are connected in parallel, then the spring constant k of combination is —m aay aes A simple pendulum is 2 avy point mass suspended by a weightless 14.63 inestensble and a perfectly flexible sting fom Rid supportaboutwhichitcan vive oa The distance between the pont of suspense, and the point of oxilaton ts called lent oe pendulum (), When the metalic bobs displaced from mean position, it executes SHIM Time pod, T= |! Vs 26. Second's pendulum, A second’s pendulum isa Pendulum whose time period is two seconds. Hs length is 993 em 27. Motion ofa tiquid ina U-tube. When a liquid of density p and contained in a U-tube upto height ‘nis depressed, it executes SHM of time period, Ta2n ft Vs 28. Motion ofa body dropped ina tunnel dug along the diameter of earth. When a body is dropped tna tunnel dug along the diameter of the earth, itexecutes SHM. If Ris radius of the earth, then its time period is Toa /K Vs 28. Motion of a body floating in a liquid. When a body made of material of density p and total vertical length L floats in a liquid of density p, such that its length his submerged in the liquid, 30. Free It a body, capable of oscillation, is slightly displaced from. its position of equilibrium and then released, it starts oscillating with a frequency of its own. Such oscillations are called free oscillations, The frequency with which a body oscillates is called natural frequency and is given by WL 22 Vm oscillations Here a body continues to oscillate with constant amplitude and fixed frequency De ati ‘The oscillations in which amplitude decreases gradually with the passage of time are called damped oscillations 31 The energy of a real oscillator decreases because a part of its mechanical energy is use ‘Sot ee NP c work against the frictional forces and islostas 32 Forced oscillations When a body os, sete if the damping force is piven PY F, =—bv, under the influence of an extemal pe, w vs the velocity of the oscillator and bisa force, not with its own natural frequency ping constant, then the displacement of the frequency of the | external periodic fo; yscillator is given by, oscillations are said to be forced oscillation — bt/2m 3. Resonant oscillations. It is a particular a ae ° forced oscillations in which fe frequency ots here wi, the angular frequency of the damped driving force is equal to the natural frequen \ lator, is given by the oscillator itself and the amplitude of o i tions is greatest. Such oscillations are called resins: | oscillations and phenomenon is called r 4ne . 34. Coupled oscillations. A system ‘of two orm " If the damping constant is small then @ =, oscillators linked together in such a way that the: where o is the angular frequency of the js mutual exchange of energy between thems | undamped oscillator. The mechanical energy E called a coupled oscillator. The oscillations of the oscillator is given by such a system are called coupled oscillations E(t) =5 kare hm, ——————— ~—