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).

Kerala Medical Entrance 2007 Questions on Simple harmonic Motion

The following two questions were included from simple harmonic motion, in KEAM (Medical) 2007 question paper:

(1) A simple pendulum has a time period T in vacuum. Its time period when it is completely immersed in a liquid of density one-eighth of the density of material of the bob is

(a) √(7/8) T (b) √(5/8) T (c) √(3/8) T (d) √(8/7) T (e) √(8/5) T

The case of a pendulum immersed in a liquid was briefly discussed in the post dated 16^th August 2006.

The mass of the bob is vρ where ‘v’ is the volume of the bob and ‘ρ’ is the density of the material of the bob. The apparent weight of the bob is v(ρ-σ)g where ‘σ’ is the density of the liquid so that the net value of the downward acceleration of the bob is v(ρ-σ)g /vρ = (1- σ/ρ)g = (1- 1/8)g = (7/8)g.

The period of oscillation of the pendulum suspended in vacuum (or, as usual, in air), is given by

T = 2π√(L/g) where 'L’ is the length and ‘g’ is the acceleration due to gravity.

When the pendulum bob is immersed in the liquid, the net acceleration is (7/8)g and the period becomes

T_liquid = 2π√(8L/7g) = √(8/7) T.

You may see the earlier post on simple pendulum containing other interesting questions, by clicking on the label ‘simple pendulum’ below this post or by making use of the search facility at the top of this page.

(2) A body of mass 20 g connected to a spring of constant K executes SHM with a frequency (5/π) Hz. The value of spring constant is

(a) 4 Nm^–1 (b) 3 Nm^–1 (c) 2 Nm^–1 (d) 5 Nm^–1 (e) 2.5 Nm^–1

This is a standard direct question. The frequency of oscillation of a spring-mass system is given by

N = (1/2π)√(K/m) where K is the spring constant and ‘m’ is the mass.

Therefore, 5/π = (1/2π)√(K/0.02), from which

K = 2 Nm^–1

A Question (MCQ) on Rolling

Here is a question which checks your understanding of basic things in angular motion and simple harmonic motion:

A solid cylinder of mass M and radius R is resting on a horizontal platform which is parallel to the XZ plane. The cylinder can rotate freely about its own axis, which is along the X-direction. The platform is given a linear simple harmonic motion of angular frequency ‘ω’ and amplitude ‘A’ in the Z-direction. If there is no slipping between the cylinder and the platform, the maximum torque acting on the cylinder is

(a) 2MR²Aω² (b) MR²Aω² (c) 2MRAω²

(d) MRAω² (e) MRAω²/2

You may be remembering the expression for maximum acceleration of a simple harmonic motion: a_max = ω²A

[ If you don’t remember the above expression, you may use the simplest form of simple harmonic motion of amplitude ‘A’ and angular frequency ‘ω’ and differentiate it twice:

z = A sin ωt (We write the displacement as ‘z’ since it is in the Z-direction).

a = d²z/dt² = –ω²A sin ωt

Therefore, the maximum acceleration is ω²A].

The maximum angular acceleration (α_max) of the cylinder is given by

α_max = a_max/R = ω²A/R.

The maximum torque (τ_max) on the cylinder is given by

τ_max = α_maxI, where ‘I’ is the moment of inertia of the cylinder about its own axis (which is equal to MR²/2).

Therefore, maximum torque τ_max = (ω²A/R) (MR²/2) = MRAω²/2

You will find an interesting MCQ on rolling at physicsplus: MCQ on Rolling Bodies

Multiple Choice Questions on Simple harmonic Motion

You will definitely find questions on simple harmonic motion in Medical and Engineering Entrance test papers as well as in GRE (Physics) question papers. You can find Multiple Choice Questions on Simple harmonic Motion with solution at physicsplus: Multiple Choice Questions on Simple Harmonic Motion . Additional questions with solution can be found here.