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Numerical Problems Based on Physics of Vibrations

Problem 1

A particle of mass 0.1 kg is situated in the potential field V = 5x2 + 10 Joule  kg-1. e down the .differential equation of motion of the particle and its frequency of oscillations.

Solution:

: Potential energy of the particle U =mass x potential= 0.1 x (5x2 + 10) Joule

 

Force F = – Du/dx= 0.1 (10x) =- x

 

But force = mass x acceleration .. F = 0.1 d2x /dt2

 

Hence equation of motion of the particle will be

0.1  d2x/dt2= -x or d2x/ dt2 +10x=0 or d2x /dt2+w2x =0

 

The above equation represents the SHM of the particle whose frequency of oscillation is

Problem 2. A particle of mass 10 g is moving along the X-axis under the force 4 x 10-2   acting towards the origin. If the initial position of rest of the particle is x = 0.1 m, write down the  differential equation of motion of particle and thus calculate ( i) the position and velocity of the particle at any instant, ( ii) the frequency of oscillation and amplitude of oscillation.

 

Solution: Given m = 10 g = 0.01 kg, force F =- 4 x 10-2 x

Hence differential equation of the motion of particle will be

and velocity of the particle at any instant t is

u = dx/dt = -0.2 sin 2t

 

(ii) Amplitude of oscillations of the particle a = 0.1 metre and frequency of oscillation

(l) n= w/2π =2/2π =0.32 Hz

 

Problem 3. A particle executing simple harmonic motion completes 1200 oscillations in 1 minute and while passing through its mean position, its velocity is 3.14 m/s.

 

(i) What is the maximum displacement of the particle from its mean position?

(ii) If at an instant t = 0, displacement is zero, obtain the equation for displacement of the  particle.

 

Hint:

(i) w = 2πn= 2π x 20 = 40π radian per second

umax = w a or 3.14 = 40πa  a = 3.14/40π = 1/40 =0.025

 

(ii) t = 0, x =a sin w  = 0.025 sin 40πt.

Problem 4. The amplitude of a particle of mass 0.1 kg executing simple harmonic motion is 0.1 m and its kinetic energy in mean position is 8 x 10-3 J. If the initial phase difference is 45°, write down the equation of motion of the particle.

 

Hint: a= 0.1 m, kinetic energy at mean position = 1/2 ma2w2 = 8 x 10 -3

 

Problem 5. A particle executes SHM of amplitude 25 cm and time period 3s. What minimum time will the particle take to move between the two points at a separation  of 12.5 cm  on either side of its mean position.

Hint: If the particle at t = 0 is at mean position, its displacement equation will be

21tt

x = 25 sin 2πt/3 . If it takes time t1 to move a distance x = 12.5 cm to one side of its mean

 

then 12.5 = 2πt1/3 :sin 2πt1/3 =sin   or  t1=1/4 s.

 

The same will be the time to move 12.5 em to the other side of its mean position, hence total time t = t1 + t2

=1/4 + ¼ =0.5 s.

Problem 6. The amplitudes of two identical simple pendulums are 2 cm and 6 cm respectively. Compare their energies of oscillations.Hint: Energy of oscillations E ∞ a2

Problem 7. The differential equation of an oscillating system is expressed as

If b << n, find the time in which: (i) the amplitude of oscillation falls to  1/e of its initial  value,

 

( ii) the energy falls to !. of its initial value.

Solution: Since b << n, hence the given equation represents the differential equation of

damped harmonic oscillator whose amplitude is a = a0 e-bt

 

(i) When the amplitude falls to 11 e times its initial amplitude a()t then

Problem 8. A particle executing damped oscillations acquires the first amplitude equal to  0 cm, starting from rest and it falls to 0.5 cm after 100 oscillations. If the time period of oscillations is 2.3 s,

find: (i) the relaxation time and (ii) the damping constant.

 

Solution: At any instant t, the amplitude of damped oscillations is

a= ao e /-t 2t  = ao e-bt

 

where ‘t is the relaxation time and b is the damping constant.

 

Given: t = 2.3 s, first amplitude at t = T / 4 is a1 = 50 cm = 0.5 cm, 201 the amplitude after

100 oscillations at t = 100 T + (T / 4) is a201 = 0. 5 cm = 0.000 m.

Thus

Problem 9. The amplitude of oscillation of a forced oscillator is expressed as

Find the value of A/A max‘ if quality factor Q = 50 and w/n = 0.99.

 

Solution: Quality factor Q = wr,2=nt  (at resonance)

t = Q/n = 50/n

Hence

At resonance (when damping is low) Amax = f t/n=50f/n2

Problem 10. The amplitude of oscillation of a forced harmonic oscillator is initially 0.02 mm, when the driving frequency is very low. It acquires a maximum value equal to 5 x 10-3 m at the frequency of driving force equal to 100Hz. Calculate (i) the quality factor of the oscillator Q,

(ii) the relaxation time-r and (iii) the half width of the resonance curve.

 

Solution: (i) Quality factor,

 

Q =    Amplitude of displacement at resonance/ Amplitude of displacement at zero or very low driving frequency

5mm/0.02 = =250

 

(ii) Since quality factor Q = wr t, where wr = 2π x 100 = 200π  rad/ s (at resonance)

 

Relaxation time t =Q/ wr = 250/200 π =0.4 second

(iii) Half width of the resonance curve

∆w /2Q = 200π /2 x 250 =0.4 π

= 1.256 rad/ s.