AIEEE Questions on Oscillations

When you prepare for any examination, it will be very useful to work out the questions which appeared in earlier examinations. Here are two questions on oscillations which appeared in AIEEE 2006 question paper:

(1) Starting from the origin a body oscillates simple harmonically with a period of 2 s. After what time will the kinetic energy be 75% of the total energy?

(1) 1/12 s

(2) 1/6 s

(3) 1/4 s

(4) 1/3 s

This simple harmonic motion can be represented by the equation,

y = A sin ωt where y is the displacement at the instant t, A is the amplitude and ω is the angular frequency.

The instantaneous velocity v of the particle is given by

v = dy/dt = Aω cosωt

The maximum velocity v_max of the particle is evidently Aω and the maximum kinetic energy which is equal to the total energy is ½ mv_max² where m is the mass of the particle. We have

½ mv² = (¾)(½)mv_max²

Therefore, ½ m (Aω cosωt)² = (¾)(½)m(Aω)² from which cosωt = (√3)/2

Therefore, ωt = π/6 so that t = π/6ω = π/(6×2π/T ) = 1/6 s since the period T is 2 s.

(2) The maximum velocity of a particle executing simple harmonic motion with amplitude 7 mm is 4.4 ms^–1. The period of oscillation is

(1) 100 s

(b) 0.01 s

(c) 10 s

(d) 0.1 s

Since the maximum velocity v_max = Aω and the period T = 2π/ω we have

T = 2πA/v_max = 2π×7×10^–3/4.4 = 0.01 s

You will find more questions (with solution) in this section here as well as here.

Kerala Engineering Entrance (2007) Questions on Oscillations

The following questions (MCQ) appeared in Kerala Engineering Entrance (2007) question paper:

(1) A particle executes S.H.M. with a time period of 16 s. At time t = 2 s, the particle crosses the mean position while at t = 4 s, its velocity is 4 ms^–1. The amplitude of motion in metre is

(a) √2 π

(b) 16√2 π

(c) 24√2 π

(d) 4/π

(e) (32√2)/π

If we start reckoning time from the instant the particle crosses the mean position (y =0 when t =0), the equation of the simple harmonic motion can be written as

y = A sin ωt, where A is the amplitude and ω is the angular frequency.

In this case, at the instant t = 2 s, we have velocity dy/dt = A ωcos ωt = 4 ms^–1.

Since the period T = 16 s, ω= 2π/T = π/8 so that the above equation becomes

A×(π/8) ×cos [(π/8)×2] = 4

Or, A×(π/8)×(1/√2) = 4, from which A = (32√2)/π

(2) In damped oscillations, the amplitude of oscillations is reduced to one-third of its initial value a₀ at the end of 100 oscillations. When the oscillator completes 200 oscillations, its amplitude must be

(a) a₀ /2

(b) a₀ / 6

(c) a₀ /12

(d) a₀ /4

(e) a₀ /9

In the case of damped oscillations, the amplitude decrease exponentially with time so that at time t, the amplitude (A) is given by

A = A₀ exp(– bt) where A₀ is the initial amplitude (at t = 0) and b is the damping constant.

If the time required for 100 oscillations is ‘t’, the time required for 200 oscillations will be 2t. Therefore we have

a₀/3 = a₀ exp(– bt) and

a = a₀ exp(– 2bt) where ‘a’ is the amplitude after 200 oscillations (at time 2t)

From the first equation, exp(– bt) = 1/3. Substituting this in the second equation,

a = a₀×(1/3)² = a₀/9.

(3) For a simple pendulum, the graph between T² and L is

(a) a straight line passing through the origin

(b) parabola

(c) circle

(d) ellipse

(e) hyperbola

The period (T) is given by T = 2 π√(L/g) with usual notations so that

T² = (4π²/g)L

This is in the form y = mx, which represents a straight line passing through the origin. So, the correct option is (a).