**Problem 1**

A particle of mass 0.1 kg is situated in the potential field V = 5x^{2} + 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 (5x^{2} + 10) Joule

Force F = – Du/dx= 0.1 (10x) =- x

But force = mass x acceleration .. F = 0.1 d^{2}x /dt2

Hence equation of motion of the particle will be

0.1 d^{2}x/dt^{2}= -x or d^{2}x/ dt^{2} +10x=0 or d^{2}x /dt^{2}+w^{2}x =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

u_{max} = 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 ma^{2}w^{2} = 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 = 2πt_{1}/3 :sin 2πt_{1}/3 =sin or t_{1}=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 = t_{1} + t_{2}

=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 ∞ a^{2}

**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 = a_{0} 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= a_{o} e /^{-t} ^{2t} = a_{o} 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.

**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 = w_{r},2=nt (at resonance)

t = Q/n = 50/n

At resonance (when damping is low) A_{max} = f t/n=50f/n^{2}

**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 = w_{r }t, where w_{r} = 2π x 100 = 200π rad/ s (at resonance)

Relaxation time t =Q/ w_{r} = 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.