Oscillation

S.K.Parida Oscillation ITER

Oscillation

Points to be remembered:

Displacement of SHM x=A(cosω0t−φ), where A= amplitude and φ=phase

Energy of SHM: ⟨E ⟩= 12m A2ω0

2

Differential equation of SHM d2 xd t 2

+ω02x=0

Differential equation of DHM d2 xd t 2

+2β dxdt

+ω02 x=0∧b

m=2 β∧k

m = ω0

2

Amplitude of the DHM A=A0 e−βt

Energy of adamped oscillator E=E0 e−2β t

Quality factorQ=ω02β

Logarithm decrement λ = log eeβT=β T

Relaxation time τ= 1β

Differential equation of forced vibration d2 xd t 2

+2β dxdt

+ω02 x= f 0cosωt

Amplitude of forced vibration for steady state A=f 0

√(ω02−ω2 )2+4 β2ω2

Phase difference between displacement and periodic force δ= tan−1( 2βωω0

2−ω2 ) Condition for amplitude resonanceω=√ω02−2 β2

Amplitude at resonance A=f 0

2 β ω0∧phaseδ=π

2

Condition for velocity resonanc e ω0=ω

1


Assignment

1. In a weakly damped oscillation, the damping force is proportional to velocity. Indicate the position at which (i) the damping force vanishes and (ii) the restoring force vanishes.

2. Give the physical significance of damping coefficient and its unit.

3. Compare the amplitude and time period of the vibration produce in free, damped and forced oscillation.

4. An under damped oscillator has a mass 0.2kg is suspended by a spring in a viscous medium whose force constant is 60 dyne/cm and damping constant 20 dyne/cm.s Show that the motion of the particle is oscillatory

5. Show that the amplitude of a weakly damped oscillation reduces to half its initial value in time t = Tln2. where T is the relaxation time.

6. Mention the conditions for under-damping, critical-damping and over-damping of a simple harmonic oscillator oscillating with angular frequency ω0 in any viscous medium of damping coefficient βand show graphical sketch of above condition.

7. A harmonic oscillator of mass 0.1kg suspended by a rigid support of spring constant 0.16N/m, oscillating under critical damping condition in any viscous medium. Calculate damping coefficient and damping constant if the oscillator moving with 3cm/sec.

8. The amplitude of a damped oscillation attains maximum value of 24cm when forced by an external periodic force Fcosωt. Find maximum amplitude of oscillation when damping co-efficient is doubled and magnitude F of the external periodic force is halved.

9. The differential equation of motion of a damped harmonic oscillating body represent by

2 ∂2 x∂ t2

+20 ∂ x∂t

+98 x=0. Find the natural frequency of the oscillating body.

10. If the displacement of an under damped oscillation representsx (t )=A e−2 t cos (10 πt−φ ). Find the constants A and φ at the mean position (x=0 ,t=0 , v=v0 ) and at the extreme position. (x=A0 , t=0 , v=0 )

11. Prove that x=(C+Dt )e−βt is the solution of the differential equation of motion for an under damped oscillator in critical damping.

12. The quality factor value for an under damped oscillator of frequency 500Hz is 5000. Calculate the

time in which (i) its amplitude reduce to 1e of its initial value.(ii) its energy reduce to

1e2

of its initial

value.

2


13. The amplitude of the oscillation of an under damped oscillator of frequency 100sec -1 fall to 1/10th

of its initial value after 1000 oscillations. Calculate (a) damping coefficient (b) logarithmic decrement (c) time in which energy falls to 1/10 of its original value.

14. A under damped oscillator oscillates with natural frequency 200Hz and its amplitude of oscillation reduced to 1/10th of its initial value after 2000cycles. Calculate

Damping coefficient Time period for damped harmonic motion Quality factor Logarithm decrement Relaxation time

15. What is forced vibration? Explain the terms ‘transient state’ and ‘steady state’ with reference to forced vibration.

16. What is resonance? What do you mean by sharpness of resonance? Under what condition it is maximum? Plot amplitude and phase versus angular frequency curve.

17. What is the phase difference between displacements of an oscillator with driving force at resonance?

18. What is the phase difference between velocity of an oscillator with driving force at resonance?

19. A forced harmonic oscillator has same displacement amplitudes at the two angular frequencies ω1rad / s∧ω2 rad /sec .Calculate the resonant frequency at which the displacement is maximum.

20. A forced harmonic oscillator has same displacement amplitudes at the two angular frequencies ω1=600 rad /s∧ω2=800 rad / s .Calculate the resonant frequency at which the displacement is maximum.

3


Solutions/Hints:

1. In a weakly damped oscillation, the damping force is proportional to velocity. Indicate the position at which (i) the damping force vanishes and (ii) the restoring force vanishes.

(Ans- at extreme position, at mean position)

2. Give the physical significance of damping coefficient and its unit. (Ans- damped motion, T -1)3. Compare the amplitude and time period of the vibration produce in free, damped and forced oscillation.

[Ans- (A , Ae− βt,f 0

√(ω02−ω2 )2+4 β2ω2

) ( 2πω0 , 2π

√ω02−β2, 2πω )

4. An under damped oscillator has a mass 0.2kg is suspended by a spring in a viscous medium whose force constant is 60 dyne/cm and damping constant 20 dyne/cm.s Show that the motion of the particle is oscillatory. (hint- show β2<ω0

2 ¿Ans:

Given: m=0.2 kg=0.2×103 g , k=60 dynecm

∧b=20 dynecm . s

ω02= km

= 600.2×103

=0.3

β2=( b2m )2

=( 202×0.2×103 )

2

=0.0025

Sinceβ2<ω02, therefore it is a oscillatory system.

5. Show that the amplitude of a weakly damped oscillation reduces to half its initial value in time t = Tln2. where T is the relaxation time. (hint : A=A0 e

−β t )Ans: A=A0 e

−β t

⇒ AA0

=e−β t

⇒ 12=e−β t

⇒2−1=e−β t

Þ ln 2−1=ln e−βt

Þ ln 2=βt

4


Þ t= ln 2β (Since T = 1/β ¿

Þ t=T ln2

6. Mention the conditions for under-damping, critical-damping and over-damping of a simple harmonic oscillator oscillating with angular frequency ω0 in any viscous medium of damping coefficient βand show graphical sketch of above condition. (see class note)

7. A harmonic oscillator of mass 0.1kg suspended by a rigid support of spring constant 0.16N/m, oscillating under critical damping condition in any viscous medium. Calculate damping coefficient and damping constant if the oscillator moving with 3cm/sec. (hint :Fd=−2m β v )

Ans: ω0=√ Km=√ 0.160.1 =1.26 rad /s

For critical damping case, β2=ω02⇒ β=ω0

Therefore damping force, Fd=−2m β v=−2mω0 v=−2×0.1×1.26×0.03=−7.6×10−3 N

Minus sign indicate that damping force is an opposing force.

8. The amplitude of a damped oscillation attains maximum value of 24cm when forced by an external periodic force Fcosωt. Find maximum amplitude of oscillation when damping co-efficient is doubled and magnitude F of the external periodic force is halved.

Ans: Amax=24 cm, β'=2β∧F '= F

2

Amax=F

2mβω0

⇒ Amax'= F '

2m β 'ω0= F/22m×2β ×ω0

= F4×2mβω0

=Amax4

=244

=6cm

9. The differential equation of motion of a damped harmonic oscillating body represent by

2 ∂2x∂ t2

+20 ∂ x∂t

+98 x=0. Find the natural frequency of the oscillating body.

Ans:

Given: 2 ∂2 x∂ t2

+20 ∂ x∂t

+98 x=0

⇒ ∂2 x∂t 2

+10 ∂ x∂ t

+49 x=0……… .. (1 )

Now, comparing the differential equation (1) with the differential equation of the damped harmonic

motion: ∂2 x∂ t 2

+2β ∂x∂ t

+ω02 x=0, we get 2 β=10∧ω0

2=49.

⇒ω0=√49=7⇒2 πν=7

⇒ ν= 72π

=1.113Hz

10. If the displacement of an under damped oscillation representsx (t )=A e−2 t cos (10 πt−φ ). Find the constants A and φ at the mean position (x=0 ,t=0 , v=v0 ) and at the extreme position. (x=A0 , t=0 , v=0 )

Ans: At the mean position, x=0 , t=0 , v=v0

5


x (t )=A e−2 t cos (10 πt−φ )

⇒0=Ae−2×0 cos (10 π×0−φ )

⇒0=Acos (−φ )

⇒cos φ=0

⇒φ=π2

V=dxdt

= ddt [A e−2 t cos (10πt−φ ) ]

⇒V=−2 A e−2 t cos (10 πt−φ )+A e−2 t ¿

⇒V 0=−2 A e−2× 0cos (10π ×0−φ )−10π A e−2×0 sin (10π ×0−φ ) ¿

⇒V 0=−2 A cos (−φ )−10π A sin (−φ)

⇒V 0=−2 A cosφ+10 π A sin φ

⇒V 0=−2 A cos π2+10π A sin π

2

⇒V 0=10Aπ

⇒ A=V 010 π

At the extreme position (x=A0 , t=0 , v=0 )

V=dxdt

= ddt [A e−2 t cos (10πt−φ ) ]

⇒0=−2 A e−2×0cos (10π ×0−φ )−10 π A e−2× 0sin (10 π×0−φ ) ¿

⇒0=−2 A cos (−φ )−10π A sin (−φ)

⇒0=−2 A cosφ+10 π A sinφ

⇒10 π A sin φ=2 A cosφ

⇒ tanφ= 210 π

⇒φ= tan−1 (0.06365 )=3.6420

x (t )=A e−2 t cos (10 πt−φ )

⇒ A0=Ae−2×0 cos (10π ×0−φ )

⇒ A0=Acos (−φ )

⇒ A0=Acos (3.6420 )=0.9979 A⇒ A≅ A0

11. Prove that x=(C+Dt )e−βt is the solution of the differential equation of motion for an under damped oscillator in critical damping. ( hint: substitute 2nd derivative, 1st derivative and x value in diff. eq. of DHM and at last show RHS goes to zero)

6


Ans: x=(C+Dt ) e−βt

⇒ dxdt

= ddt

(C+Dt ) e−βt=−Cβ e− βt+De−βt−Dtβe− βt

⇒ d2 xd t 2

= ddt. dxdt

= ddt

[−Cβ e− βt+De−βt−Dtβ e−βt ]=C β2 e−βt−Dβ e−βt+Dt β2e−βt−Dβ e−βt

We know the differential equation of DHM is d2 xd t 2

+2β dxdt

+ω02 x=0 ……………………. (1)

The LHS of the eq(1),

¿C β2 e−βt−Dβ e−βt+Dt β2e− βt−Dβ e−βt+2 β (−Cβ e−βt+De−βt−Dtβ e−βt )+ω02 (C+Dt ) e−βt

=

C β2 e−βt−Dβ e−βt+Dt β2 e−βt−Dβ e−βt−2C β2e−βt+2Dβ e− βt−2Dt β2 e−βt+C ω02 e−βt+Dtω0

2e−βt

=

C β2 e−βt+C ω02 e−βt−2C β2e−βt−Dβ e− βt−Dβ e−βt+2Dβ e−βt+Dt β2e−βt+Dt ω0

2e−βt−2Dt β2e−βt

Under the critical damping condition ω02=β2

= C β2 e−βt+C β2e−βt−2C β2e− βt+Dt β2 e−βt+Dt β2e−βt−2Dt β2e−βt=0

12. The quality factor value for an under damped oscillator of frequency 500Hz is 5000. Calculate the time

in which (i) its energy reduces to 1e of its initial value (ii) how many oscillations the oscillator will make

in this time.Ans:

Given: ν=500Hz∧Q=5000

ω0=2πν=2×3.142×500=3142cyclesec

Q=ω02 β

⇒5000=31422β

⇒ β= 314210000

=0.3142 s−1

For case (i) we know, E=E0 e−2β t

Þ EE0

=e−2β t

7


Þ 1e=e−2× 0.3142× t (From the value ofβ)

⇒−1=−2×0.3142× t

Þ t= 10.6284

=1.59 s

For case (ii) the energy is reduced 1/e to the value in 1.59 second which is equal to the relaxation time

τ . The number of oscillations made by the oscillator in time t will be n=( ω02π ). τn=2 βQ

2π. τ ¿

Qπ

= 10003.142

=318 (Since Q=ω02 β

)

13. A under damped oscillator oscillates with natural frequency 200Hz and its amplitude of oscillation reduced to 1/10th of its initial value after 2000cycles. Calculate

Damping coefficient Time period for damped harmonic motion Quality factor Logarithm decrement Relaxation time

Ans: Given ω0=2πν=2×3.142×200=125.68 cycle /sec

Time = t= 2000cycles = 2000T, A=A010

, T=1ν= 1200

sec

We know A=A0e−β t

Þ AA0

=e−β t

Þ 110

=e−β [2000. T ] (From given condition of question)

log e10−1=logee

− β [2000 .T ]

Þ −2.303× log10 = −β .2000T

Þ 2.303 = β .2000 . 1200

β=2.30310

=0.2303 sec−1

Time period for damped harmonic motion = T= 2π

√ω02−β2 = 2X 3.142

√(125.68)2−(0.2303)2=0.000397 s

8


Quality factor= Q=ω02 β

=125.682 X0.2303

=272.86

Logarithmic decrement: λ ¿ βT = 0.2303×( 1200 )=0.00115

Relaxation timeτ=1β= 10.2303

=¿ 4.34 s

14. The amplitude of the oscillation of an under damped oscillator of frequency 100sec -1 fall to 1/10th of its initial value after 1000 oscillations. Calculate (a) damping coefficient (b) logarithmic decrement (c) time in which energy falls to 1/10 of its original value.

Ans: Time = t= 1000cycles = 1000T,A=A010

, T=1ν= 1100

sec

We know A=A0 e−β t

Þ AA0

=e−β t

Þ 110

=e−β [1000 . T ] (From given condition of question)

log e10−1=logee

− β [1000 . T ]

Þ −2.303× log10 = −β .1000T

Þ 2.303 = β .2000 . 1100

β=2.30320

=0.115 sec−1

Logarithmic decrement: λ ¿ βT = 0.115×( 1100 )=0.00115 We know, E=E0 e

−2β t

Þ EE0

=e−2β t

Þ 110

=e−2×0.115× t (From the value ofβ∧EE0

= 110 )

log e10−1=logee

−0.23 t

Þ −2.303× log10 = −0.23 t

Þ t = 2.3030.23

=10.01 s

9


15. What is forced vibration? Explain the terms ‘transient state’ and ‘steady state’ with reference to forced vibration. ( hint: see the class note)

16. What is resonance? What do you mean by sharpness of resonance? Under what condition it is maximum? Plot amplitude and phase versus angular frequency curve.

Ans: The rapidity in which the amplitude falls as the forcing frequency defers from resonance on either side of the maximum value of amplitude is called sharpness of resonance. Condition for width of the resonance , Pav=12 (Pav )max ------------------------ (1)Pav=βm A

2ω2 and Amax= f 02βω

(Pav )max=βm.( f 02 βω )

2

.ω2=m f 0

2

4 β

Substituting above value in eq.(1), we getβm A2ω2=1

2×m f 0

2

4 β

βm×f 02ω2

(ω02−ω2 )2+4 β2ω2

=m f 0

2

8 β

⇒8 β2ω2=(ω02−ω2 )2+4 β2ω2

⇒ω0−ω=±2βωω0+ω

∴ω0−ω1= 2βωω0+ω

-------------------------- (2)ω0−ω2=

−2βωω0+ω

---------------------------- (3) ( reference the figure) Eq.(2) - eq.(3), we get ω2−ω1= 4 βω

ω0+ω

⇒ω2−ω1= 4 β

1+ω0ω

, at the resonance, ω0=ω

10


Therefore ω2−ω1= 4 β1+1

=2 β

Hence, the width of the resonance ∆ ω=(ω¿¿2−ω1) isdepends uponthe dampingcoefficients 2 β .¿ Large the damping, amplitude fall will be smaller. Therefore the sharpness of resonance will be small. So the curve is very flat. But for weak damping, resonance will be very sharp.

17. What is the phase difference between displacements of an oscillator with driving force at resonance? (Ans-900)

18. What is the phase difference between velocities of an oscillator with driving force at resonance? (Ans-00)19. A forced harmonic oscillator has same displacement amplitudes at the two angular frequencies

ω1rad / s∧ω2 rad /sec .Calculate the resonant frequency at which the displacement is maximum.

Ans: here given condition of the question, A1=A2

⇒ f 0

√(ω02−ω12 )2+4 β2ω1

2=

f 0

√ (ω02−ω22)2+4 β2ω2

2

Squaring on both side, we get

⇒ (ω02−ω12 )2+4 β2ω1

2=(ω02−ω22 )2+4 β2ω2

2

⇒ (ω02−ω12 )2−(ω02−ω22 )

2=4 β2ω2

2−4 β2ω12

⇒ (ω02−ω12+ω02−ω22 ) (ω02−ω12−ω02+ω22 )=4 β2 (ω22−ω12 )

⇒ (2ω02−ω12−ω22 ) (ω22−ω12 )=4 β2 (ω22−ω12 )

⇒ (2ω02−ω12−ω22 )=4 β2

⇒2ω02−4 β2=(ω12+ω22)

⇒2 (ω02−2 β2)=ω12+ω22

Where resonant frequency, ωr=ω02−2 β2

⇒2ωr2=ω1

2+ω22

⇒ωr=√ω12+ω222

20. A forced harmonic oscillator has same displacement amplitudes at the two angular frequencies ω1=600 rad /s∧ω2=800 rad / s .Calculate the resonant frequency at which the displacement is

maximum. (hint: ωr=√ω12+ω222)

11


Ans: ωr=√(600)2+(800)2

2=707.1 rad /s

12

Documents

Oscillation