Paper-II Chapter- Damped vibratio Free vibratios: Whe a body cotiues to oscillate with its ow characteristics frequecy. Such oscillatios are kow as free or atural vibratios of the body. Ideally, the body vibrates with its atural frequecy for a idefiite time with a costat amplitude. Damped vibratios: I practice, the amplitude of vibratios decays with time ad fially the body comes to rest at its mea positio. So, we may coclude that there is a dampig force o the vibratig body ad this may be due to viscosity of the medium or other frictioal forces. Such vibratios of decayig amplitude are referred to as resisted or damped vibratios. The motio of a simple pedulum is a example of a damped simple harmoic motio, because the amplitude gradually decreases with time ad the bob fially comes to rest. Aalytical treatmet of damped vibratios: Let x be the displacemet of a particle of mass m performig a damped motio from its iitial positio of rest. The followig forces act o the body. i Restorig force F 1 : The restorig force F 1 is proportioal to the displacemet x ad teds to brig the particle back to its iitial positio. i.e. F 1 x 1
where s is called the stiffess costat. F 1 = sx 1 ii Retardig or resistig force F : This force is caused by frictio. This force is proportioal to the istataeous velocity of the particle. F v F 1 = kv = k dx dt where k is the restorig force per uit velocity. The -ve sig idicates that the force opposes the motio. The system is subjected to viscous frictio. So, the et force actig o the bob is F = F 1 + F F = sx k dx dt From Newto s d law of motio, we get So, from 3 we get where b = k m ad ω = s m. 3 F = m d x dt 4 m d x dx = sx k dt dt d x dt = s m x k dx m dt d x dt + k dx m dt + s m x = 0 d x dt + bdx dt + ω 0x = 0 5 This is the differetial equatio of damped harmoic motio. Let auxiliary solutio of the equatio 5 is x = Ae mt
So we get from 5 m + bm + ω = 0 m = b ± b ω So, we ca write m 1 = b + b ω m = b b ω So, the solutio of 4 is x = A 1 e m 1t + A e m t x = A 1 e b+ b ω t + A e b b ω t x = e bt A 1 e b ω t + A e b ω t x = e bt A 1 e b ω t + A e b ω t x = e bt A 1 e b ω t + A e b ω t 6 Where A 1 ad A are costat to be determied from the iitial coditio. Let b = b ω. From 6 we get q = e bt A 1 e b t + A e b t 7 Case I: Over damped motio b > ω 0 : Whe dampig is very large, i.e. b > ω 0 Now, at t = 0, x =, from 7 we get At t = 0, dx dt = v 0, from 7 we get dx dt = be bt A 1 e b t + A e b t = A 1 + A A 1 + A = 8 t=0 + e bt b A 1 e b t b A e b t t=0 3
Solvig 8 ad 9, we get A 1 = A = So, we get the fial solutio x = e bt x = 1 e bt v 0 = ba 1 + A + b A 1 b A v 0 = b + b A 1 A A 1 A = b + v 0 b 9 1 + b + v 0 b 1 b + v 0 b 1 + b + v 0 = = b ω e b ω t + 1 + b + v 0 x 0 e b ω t + b ω The variatio of x with time t is show i fig. 1 + b + v 0 x 0 b ω 1 b + v 0 x 0 b ω 1 b + v 0 x 0 e b ω t b ω 1 b + v 0 b ω e b ω t 10 11 The motio of the particle is clearly aperiodic. The displacemet falls off ad asymptotically approaches zero with icreasig time. This is the over damped motio ad is exhibited by a pedulum movig i a highly viscous liquid or by a aperiodic movig coil galvaometer. Case II: For b < ω the b ω = ω b = j ω b = jω. From 7 we get 1 x = e bt 1 + b + v 0 e jω t + 1 b + v 0 e jω t jω jω 4
e bt x = e jω t + e t jω + b + v 0 e jω t e jω t jω e x = e bt jω t + e jω t + b + v 0 e jω t e jω t ω j x = e bt cosω t + b + v 0 siω t 1 ω Let AcosΘ = 1 ad AsiΘ = b+ v 0, So ω ad A = 1 + b ω = 1 + b + v 0 ω b Θ = ta 1 b + v 0 ω b So we get from 1 x = e bt A cosω tcosθ + siθsiω t cos x = Ae bt ω b t Θ cos x = Re bt ω b t Θ 13 where R = A Eq.13 shows that the motio is oscillatory with a agular frequecy ω = ω0 b ad a amplitude Re bt. Sice the amplitude decays expoetially with time, the motio is damped oscillatory. The variatio of x with time t is show i fig. Here, the oscillatios soo die dow as the amplitude is proportioal to e bt. 5
If the particle is iitially displaced to ad the released, v 0 = 0.,The A = ω 0 ω 0 b So we get x = ω 0 e ω bt cos ω b t Θ 0 b Θ = ta 1 b ω b Case II: Critical dampig b ω 0 Let b ω 0, so we take ω b = δ. So we get from 11 x = x 0 e bt 1 + b + v 0 x 0 e b ω t + b ω x = x 0 e bt 1 + b + v 0 x 0 1 + δt + b ω x = e 1 bt + b + v 0 1 b + v 0 b ω e b ω t 1 b + v 0 b ω 1 δt This shows that the motio is aperiodic, but x approaches zero quicker tha the over damped case. The motio is ow said to be critically damped, the variatio of x with t beig displayed by fig. t Q. Derive the differetial equatio of damped motio from cosideratio of the eergy of the system. We have kietic eergy of the system ad the potetial eergy E K.E. = 1 dx m dt E P.E. = 1 sx 6
Let the particle be described a elemet of displacemet δx, the the loss of kietic ad potetial eergy of the particle will be equal to the work doe agaist the frictioal force. Hece But dx dt δ E K.E. + E P.E. = dw = k dx dt δx 1 dx δ m dt dx dt 0 for all values of t. Hece, 1 + sx d x dt + bdx dt + ω 0x d x dt + bdx dt + ω 0x = 0 = k dx dt δx = 0 Forced Vibratio: A vibratig system gradually loses its amplitude sice eergy is dissipated due to frictioal forces. To maitai the system i vibratio, eergy must be supplied from outside. If a exteral periodic force is applied to the vibratig system, the system teds to vibrate with its ow atural frequecy. But the applied drivig force tries to impress its ow frequecy o the vibratig system. Iitially, the system vibrates with both the frequecies. The atural vibratios die out i course of time due to the prevailig resistig forces ad the system fially i the steady state, vibrates with the frequecy of the drivig force with a costat amplitude. Such vibratio where the system oscillates with a frequecy the same as that of a exterally impressed periodic force is kow as the forced vibratio of the system. Aalytical treatmet of forced vibratio: Let a particle of mass m, capable of executig a damped simple harmoic motio, be subjected to a exteral simple harmoic forced of costat amplitude ad frequecy. Let x is the displacemet of the particle from its mea positio at time t. 7
The forces actig o the particle are the followig: i The restorig force F s = sx tedig to brig the particle back to its mea positio, s beig the stiffess factor. ii The resistig force F r = k dx, k beig the resistig force per uit velocity. dt iii The drivig periodic force F d = F cosωt, where F is the amplitude ad ω is the agular frequecy of the force. The et force i the +ve x-directio is F = F d + F r + F s F = F cosωt k dx dt sx From Newto s laws of motio, we get m d x dt d x dt dx = F cosωt k dt sx = F m cosωt k dx m dt s m x d x dt + bdx dt + ω 0x = fcosωt 1 where b = k/m ad ω 0 = s/m ad f = F m. The quatity ω 0 is the udamped atural agular frequecy of the particle ad b is the decay costat. For the atural motio of the particle, the differetial equatio is d x dt + bdx dt + ω 0x = 0 So the complemetary fuctio is cos x 1 = Re bt ω b t Θ 8
where R ad Θ are arbitrary costats to be determied from the iitial coditios. Let x r is the displacemet for the drivig force F cosωt. So the equatio of motio for this case d x dt + bdx dt + ω 0x = fcosωt 3 If x i is the displacemet for the drivig force F siωt. So the equatio of motio for this case d x dt + bdx dt + ω 0x = fcosωt 4 Multiplyig 4 by i 1 ad addig 3, we get Let x r is the displacemet for the drivig force F cosωt. So the equatio of motio for this case d x r + ix i + b dx r + ix i + ω dt dt 0x r + ix i = fcosωt + ifsiωt d X dt + bdx dt + ω 0X = fe ωt 5 where X = x r + ix i. Let the particular itegral of 5 X = X 0 e iωt where X 0 is a complex quatity. Substitutig i 5, we get X 0 ω + ibω + ω 0 e iωt = fe iωt Let ad f X 0 = ω0 ω + ibω f ω0 ω ibω X 0 = 6 ω0 ω + bω acosφ = ω 0 ω asiφ = bω 9
Hece, ad From 6 we get] So, the solutio a = ω 0 ω + 4b ω taφ = bω ω 0 ω f acosφ iasiφ X 0 = ω0 ω + 4b ω fa cosφ isiφ X 0 = ω0 ω + 4b ω X 0 = X = fe iφ ω 0 ω + 4b ω X = X 0 e iωt fe iφ ω 0 ω + 4b ω eiωt x r + ix i = X = f ω 0 ω + 4b ω eiωt φ f cosωt φ + isiωt φ ω0 ω + 4b ω So for the drivig force for F cosωt, the particular itegral is x r = x r = x r = f cosωt φ ω 0 ω + 4b ω F cosωt φ m ω 0 ω + 4b ω F cosωt φ mω 0 mω + 4b m ω x r = F cosωt φ k ω + s mω 10
Similarly, for the drivig force F siωt, we get So, the geeral solutio x i = F siωt φ k ω + s mω x = x 1 + x r cos x = Re bt F ω b t Θ + cosωt φ k ω + s mω The first term o the right had side gives the atural damped simple harmoic motio of a particle. The secod term gives the forced vibratio. Iitially, both the terms are operative ad their superpositio gives a irregular motio. After a few time costats, the atural vibratio dies out sice its amplitude falls off expoetially with time with the decay costat b. Thus, fially oly the forced vibratio give by the secod term persists. This is the steady state motio. SO, i steady state x = F cosωt φ k ω + s mω Amplitude or displacemet resoace: Whe the displacemet amplitude is a maximum for some frequecy of the driver, we have the pheomeo of amplitude or displacemet resoace. We have at the steady-state displacemet amplitude of forced vibratio as A = F k ω + mω s 1 The amplitude A is maximum for that value of ω for which the deomiator o the right-had side of 1 is a miimum. So, we get k ω + s mω = miimum d k ω + s mω = 0 dω 11
k ω + 4mω sωm = 0 k + mω sm = 0 ω = s m k m ω = ω 0 b Substitutig the value of ω i 1, we get the maximum displacemet amplitude A m = F/m b ω0 b The variatio of the displacemet amplitude with the agular frequecy of the driver differet values of the decay costat is show i fig. Velocity resoace: Whe the velocity amplitude attais a maximum for a certai frequecy of the drivig force, we have velocity resoace. If the forcig fuctio is F cosωt, the steady-state displacemet x = F siωt φ k ω + s mω So, the velocity v = dx dt = F ω cosωt φ k + mω s ω 1
whe ωm s = 0, the v attais its maximum value ω Hece velocity resoace occurs whe v m = F k ωm s ω = 0 ωm = s ω ω = ω 0 i.e. whe the forcig frequecy equals the udamped atural frequecy of the forced system. Plot of the velocity amplitude of the forcig system versus agular frequecy of the driver for differe values of the decay costat is show i fig. Power relatios i Forced vibratio: We have work doe W = F cosωtdx I the steady state, the power of the driver i forced vibratio is F = P = F cosωt dx dt F ω cosωtcosωt φ k ω + s mω F ω F = cos ωtcosφ + siωtcosωtsiφ k ω + s mω F ω P = cos ωtcosφ + 1 k ω + s mω siωtsiφ The average power over a complete cycle is = 1 T T 0 P dt 13
= 1 T T 0 F ω cos ωtcosφ + 1 k ω + s mω siωtsiφ dt = 1 F ω T T k ω + s mω cosφ + 0 = 1 F ω k ω + s mω cosφ Agai So, average power cosφ = = k k ω + s mω F ωk k ω + s mω At resoace we have k + s mω = k. So the average power at resoace res = F k Sharpess of resoace: We have average power = F k 1 k + mω s ω At resoace i.e. at ω = ω 0 = s m max = F If is plotted agaist ω, clearly, will attai its maximum value. ad will fall off o either side of this frequecy as show i fig. k 14
Hece, resoace is sharp whe dampig k is small. For large dampig, the resoace is broad or flat which is show i fig. The sharpess of resoace gives the rapidity with which the average power drops off as ω differs from its value at resoace. Hece, higher sharpess of resoace is for low dampig. Average power at resoace Dividig 1 ad, we get res = F k res res res res = = = = k + k mω s ω k k + mω sω k k + m ω ω 0 ω 4b + ω 0 15 4b ω ω 0 ω 0 ω
where = ω ω 0 ω 0 ω. res = 4b 4b + ω 0 At ω = ω 0, i.e. at = 0, Bellow or above resoace res 0 = 1 ad res The sharpess of resoace is quatitatively defied as the reciprocal of at which < 1 res = 1 The above fig. shows that for a give value of k or b, there are two values of ω, amely, ω 1 ad ω for which the average power of is half its value at resoace. These frequecies are called the half-power frequecies. At the half-power frequecies, 16
we get res = 4b 4b + ω 0 1 = 4b 4b + ω 0 This gives the sharpess of resoace as ω 0 = 4b ω 0 = ±b S r = 1 = ω 0 b So, two half-power frequecies are ω ω 0 ω = ±b ω bω ω 0 = 0 ω 1 = b + ω 0 b ad So, we get ω = b + ω 0 + b ω ω 1 = b Hece, the sharpess of resoace S r = 1 = ω 0 b = ω 0 ω ω 1 Agai ω 1 ω = b + ω0 b b + ω0 b ω 1 ω = b + ω0 b ω 1 ω = ω 0 17
Doppler Effect: The pitch of a ote i.e. its frequecy as perceived by a listeer, appears to chage whe there is a relative motio betwee the source of soud ad the listeer. The apparet iequality betwee the emitted ad the perceived frequecies is referred to as Doppler effect. i Source movig, observer at rest: Let O be the observer at rest. S be the source movig with a velocity V s alog SO. The waves of frequecy propagate towards the observer with a velocity V, if the source is statioary. So, the origial wavelegth of soud λ = V Now, i t = 1 sec the wave travels the distace SO = V t = V Let the source S moves to S i t = 1 sec., SS = V s t = V s So, the waves of frequecy will occupy the distace i t = 1 sec S O = V V s 18
So, the chaged wavelegth λ = S O = V V s As the velocity of soud remai costat, the apparet frequecy of the ote is = V λ = Hece, the apparet chage i frequecy = = So, the fractioal chage i frequecy V V V s 1 V 1 V V s V s V V s = V s V V s Case I: If the source moves away from the observer, V s is -ve, so apparet frequecy of the ote = V V + V s Case II: Whe the wid blows with velocity V w i the directio of soud, the soud velocity is icreased to V + V w. So, we get the apparet frequecy from 1 = V + V w V + V w V s Case II: Whe the wid blows with velocity V w i the opposite directio of soud, the soud velocity is decreased to V V w. So, we get the apparet frequecy from 1 = V V w V V w V s II. Observer movig, source at rest: Let the observer O moves with velocity V O away from the source. 19
If the observer is at rest, the the waves crossig O i t = 1 sec would occupy a legth OA = V. So, the origial wavelegth of soud λ = OA = V Now, i t = 1 sec the observer moves to O. Hece OO = V 0 So, i t = 1 sec. the observer receives the waves occupyig the legth So, the apparet frequecy of the ote is O A = V V O = V V O λ = V V O V Case I: If the observer approaches the source, V O is -ve, so apparet frequecy of the ote = V + V O V Case II: Whe the wid blows with velocity V w i the directio of soud, the soud velocity is icreased to V + V w. So, we get the apparet frequecy from 1 = V + V w V O V + V w Case II: Whe the wid blows with velocity V w i the opposite directio of soud, the soud velocity is decreased to V V w. So, we get the apparet frequecy from 1 = V V w V O V V w III Both source ad observer movig: Whe the observer is at rest ad the source moves, the apparet frequecy is 1 = V V V s 1 0
Whe the source of emittig frequecy 1 is statioary, but the observer is movig towards the source, the apparet frequecy is = V + V O 1 V = V + V O V V s 3 Case I: Whe the wid blows with velocity V w i the directio of soud, the soud velocity is icreased to V + V w. So, we get the apparet frequecy from 3 = V + V w + V O V + V w V s 4 1