IIT-JEE 2011 Questions (MCQ) on Simple Harmonic Motion

“Do not worry about your problems with mathematics, I assure you mine are far greater.”

– Albert Einstein

The following two questions on simple harmonic motion were included in the IIT-JEE 2011 question paper. Both questions are simple for those who have mastered the fundamental principles:

(1) A wooden block performs SHM on a frictionless surface with frequency, ν₀. The block carries a charge +Q on its surface. If now a uniform electric field E is switched on as shown, then the SHM of the block will be

(A) of the same frequency and with shifted mean position

(B) of the same frequency and with the same mean position

(C) of changed frequency and with shifted mean position

(D) of changed frequency and with the same mean position

When the uniform electric field is switched on, the positively charged block experiences a constant force in the direction of the electric field and so the mean position of the block is shifted. But the frequency (ν) of oscillation of the block is unchanged since it depends only on the mass m of the block and the force constant k of the spring, in accordance with the relation

ν = (1/2π) [√(k/m)]

The correct option is (A).

(2) A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x₁(t) = A sin ωt and x₂(t) = A sin(ωt + 2π/3). Adding a third sinusoidal displacement x₃(t) = B sin(ωt + φ) brings the mass to a complete rest. The value of B and φ are

(A) √2 A, 3π/4

(B) A, 4π/3

(C) √3 A, 5π/6

(D) A, π/3

The particle in the above question is subjected to two simultaneous simple harmonic motions (SHM) of the same frequency and amplitude in the same direction. the resultant motion of the particle is simple harmonic as is evident by adding x₁(t) and x₂(t):

x₁(t) + x₂(t) = A sin ωt + A sin(ωt + 2π/3) = A sin(ωt +π/3)

The amplitude of the resultant simple harmonic motion is A itself but the initial phase of the motion is now π/3.

As the particle remains at rest on adding the third simple harmonic motion, x₃(t) = B sin(ωt + φ) the amplitude (B) of the third SHM must be A itself, but it must be 180º (or, π radian) out of phase. In other words, the initial phase (φ) of the third simple harmonic motion must be π/3 + π = 4π/3

The correct option is (B).

You can find a useful post on simple harmonic motion here.

Questions (MCQ) on Oscillations of a Spring-mass System

The following questions involving a spring-mass system are meant for checking whether you have a thorough understanding of the simple harmonic oscillations in which the restoring force is supplied by a spring:

(1) The frequency of vertical oscillations of a mass suspended at the end of a light spring is n. If the system is taken to a location where the acceleration due to gravity is reduced by 0.1%, the frequency of oscillation will be

(a) 1.01 n

(b) 0.99 n

(c) 1.001 n

(d) 0.999 n

(e) n

A spring-mass system (unlike the simple pendulum) does not require a gravitational force for oscillations since the restoring force required for oscillations is supplied by the elastic forces in the spring.

[Note that the period (T) of oscillations is given by T = 2π√(m/k) where m is the mass attached to the spring and k is the spring constant].

Therefore the change in ‘g’ does not affect the frequency and the correct option is (e).

(2) One end of a light spring is fixed to the ceiling and a mass M is suspended at the other end. When an additional mass m is attached to the mass M, the additional extension in the spring is e. The period of vertical oscillation of the spring-mass system now is

(a) 2π√[(M+m)e/mg]

(b) 2π√(me/mg)

(c) 2π√[(M+m) /mge]

(d) 2π√[me/(M+m)g]

(e) 2π√(M/mge)

The period (T) of oscillations is given by T = 2π√[(M+m)/k] where k is the spring constant.

Since an additional weight mg attached to the spring produces an additional extension e, the spring constant k = mg/e.

Therefore, period of oscillations T = 2π√[(M+m)/(mg/e)] = 2π√[(M+m)e/mg], as given in option (a).

(3) The period of vertical oscillations of a mass M suspended using a light spring of spring constant k is T. The same spring is cut into three equal parts and they are used in parallel to suspend the mass M as shown in the adjoining figure. What is the new period of oscillations?

(a) T

(b) 3T

(c) 9T

(d) T/9

(e) T/3

The original period of oscillation T is given by

T = 2π√(M/k)

When the spring is cut into three equal parts, each piece has spring constant 3k.

[Since the length of each piece is reduced by a factor three, the extension for a given applied force will be reduced by a factor three so that the spring constant (which is the ratio of force to extension) will become three times].

Since the three pieces are connected in parallel the effective spring constant of the combination is 3k+3k+3k = 9k. The new period of oscillations T₁ is given by

T₁ = 2π√(M/9k) = T/3 since T = 2π√(M/k)

You will find some multiple choice questions with solution in this section here.

Two Questions on Oscillations

Today we will discuss two multiple choice questions on simple harmonic motion:
(1) A simple pendulum of period 2 s has a small bob of mass 50 g. The amplitude of oscillation of the bob is 10 cm and it is at a height of 45 cm from the ground in its mean position. While oscillating, the string breaks just when the bob is in its mean position. The horizontal distance R `from the mean position where the bob will strike the ground is nearly

(a) 35.2 cm (b) 23 cm (c) 15.3 cm

(d)12.4 cm (e) 9.4 cm

The angular frequency (ω) of oscillation of the pendulum is given by

ω = 2π/T = 2π/2 = π rad/s.

The bob has maximum velocity (v_max) in the mean position and is given by

v_max = ωA where A is the amplitude.

Therefore, v_max = π × 0.1 = 0.1 π ms^–1

On getting detached from the string, the bob moves like a projectile shot horizontally from a height of 0.45 m with a velocity of 0.1 π ms^–1. Its time of flight (t) is obtained from the vertical displacement of 0.45 m:

0.45 = 0 + ½ gt². (Note that the initial vertical velocity is zero and the vertical acceleration is g, which we may take as 10 ms^–2). This gives t² = 0.09 so that t = 0.3 s.

The horizontal distance covered by the bob during this time is 0.1 π × 0.3 = 0.094 m = 9.4 cm.

[Note that the mass of the bob does not come into the picture and it just serves as a distraction].

Now, consider the following MCQ :

A large horizontal surface moves up and down simple harmonically with an amplitude of 1 cm. If a mass of 3 kg (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of the SHM will be

(a) 5 Hz (b) 2 Hz (c) 8 Hz (d) 10Hz (e) 15 Hz

The mass will remain in contact with the surface if the maximum acceleration produced in the simple harmonic motion does not exceed the acceleration due to gravity (g).

Therefore, we have

ω²A = g where ω is the angular frequency of the SHM.

From this ω = √(g/A) = √(10/ 0.01) = 10×√10.

The maximum frequency (linear) of oscillations is therefore given by

n = ω/2π = (10×√10)/2π = 5 Hz.

[Note that the mass of the body placed on the surface does not come into the picture and it just serves as a distraction].

MCQ on Oscillation of Magnets

The period (T) of angular oscillations of a magnet suspended freely in a magnetic field of flux density ‘B’ is given by
T = 2π √(I/ mB) where ‘I’ is the moment of inertia of the magnet about the axis of the angular motion and ‘m’ is the magnetic dipole moment. You will find questions based on this relation in your entrance examination question papers. Here is a question which is meant also for checking your grasp of the moment of inertia and the vector property of the magnetic dipole moment:
Two identical magnets are placed one above the other and tied together so that their like poles are in contact. The period of oscillation of this combination (on suspending horizontally using torsionless suspension fibre) in a horizontal magnetic field is 2 s. What will be the period of oscillation if the magnets are placed one above the other such that they are mutually perpendicular and bisect each other?
(a) 2 s (b) √2 s (c) 2√2 s (d) 2^¼ s (e) 2×2^¼ s

The period of oscillation is given by T = 2π √(I/ mB) with usual notations.
In the both cases, the moment of inertia is twice that of one magnet. In the first case, the net dipole moment is twice that of one magnet since the like poles are pointing in the same direction. In the second case, the net dipole moment is √2 times the moment of one magnet since the moment vectors at right angles get added.
The equations for the period in the two cases are therefore
T1 = 2π √(2I/ 2mB) = 2π √(I/ mB) = 2 s (as given in the question) and
T2 = 2π √(2I/ √2 mB) = 2π √(√2 I/ mB) = T1 √(√2) = 2×2^¼ s.
The following MCQ appeared in EAMCET (Med) A.P.2003 question paper:
The period of oscillation of a magnet at a place is 4 seconds. When it is remagnetised so that the pole strength becomes 4 times the initial value, the period of oscillation in seconds is
(a) ½ (b) 1 (c) 2 (d) 4

The period of oscillation is given by T = 2π √(I/ mB) with usual notations. The period is therefore inversely proportional to the square root of the dipole molent ‘m’. When the pole strength is made 4 times the initial value, the dipole moment is made 4 times the initial value so that the period becomes half the initial value. The correct option is 2 seconds.

Additional Multiple Choice Questions on Simple Harmonic Motion

The following post is in continuation of the post ‘Multiple Choice Questions on Simple Harmonic Motion’, dated 19th October 2006:

Let us consider the following MCQ which appeared in Kerala Engineering Entrance 2005 question paper:

A particle starts SHM from its mean position. Its amplitude is ‘a’ and total energy is ‘E’. At one instant its kinetic energy is 3E/4. Its displacement at that instant is
(a) a/√2 (b) a/2 (c) a/√(3/2) (d) a/√3 (e) a

The equation of the simple harmonic motion is y = A sinωt with usual notations. The total energy is (½) m ω²a² and the kinetic energy is (½)m ω²(a² –y²).
Therefore we have (½) m ω²a² = E and
(½)m ω²( a² –y²) = 3E/4.
Dividing the second equation by the first, 1 – y²/a² = ¾, from which y = a/2.
The following question will be interesting to you:

Two discs A and B of masses m and 2m respectively are connected by a light spring and placed on a table (Fig.) so that the disc B is on the table and the disc A is resting on the spring. If the frequency of oscillation of the spring on slightly depressing and releasing the disc A, is ‘f’, the frequency of oscillation on suddenly removing the table is (a) f/2 (b) f/√2 (c) f (d) √(2/3) f (e) √(3/2) f
When disc B is resting on the table, disc A functions as the oscillating mass in a spring-mass system and the period of oscillation, as usual, is given by T = 2π√(m/k) where ‘k’ is the spring constant. The frequency of oscillation is f = (1/2π)√(k/m).
When the table is removed, the system falls freely and both discs are free to oscillate. The reduced mass (μ) of the vibrating system is given by μ = m×2m/(m+2m) = (2/3)m.
The frequency of vibration in this case is obtained by replacing ‘m’ in the above expression for frequency by (2/3)m so that the new frequency is (1/2π)√[k/(2/3)m] = (1/2π)√(k/m) ×√(3/2) = f ×√(3/2) [Option (e)].
Now, consider the following question:
The kinetic energy and potential energy of a simple harmonic motion (of amplitude ‘A’ ) will be equal when the displacement is
(a) A/√2 (b) A√2 (c) A/2 (d) A/3 (e) A/√3
Equating the kinetic energy to the potential energy, we have ½ m ω²( A² –y²) = ½ m ω²y², from which ( A² –y²) = y². Therefore, y = A/√2.
Consider the following MCQ: Two light springs of force constants k₁ and k₂, loaded with equal masses undergo vertical oscillations. If the maximum velocities of the masses are the same, the ratio of the amplitudes of oscillations is
(a) k₁ /k₂ (b) 1 (c) k₁ ² /k₂ ² (d) k₂ /k₁ (e) √(k₂ /k₁ )
If A₁ and A₂ are the amplitudes of oscillations of the first and the second system respectively and ω1 and ω2 their angular frequencies, we have ω₁ A₁ = ω₂ A₂ , since the maximum velocities are equal. Therefore, A₁ /A₂ = ω₂ /ω₁ .
But ω = √(k/m) for linear simple harmonic motion so that A₁ /A₂ = √(k₂ /m₂ )/ √(k₁ /m₁ ) = √(k₂ /k₁ ) since the masses are equal.
[Even if you don’t remember the relation, ω = √(k/m), you can work out this since the maximum potential energy (and maximum kinetic energy) of an oscillating spring is ½kA² . Since the maximum velocities are equal (as specified in the question), the maximum kinetic energies are equal so that we have, ½k₁ A₁ ² =½k₂ A₂ ² , from which A₁ /A₂ = √(k₂ /k₁ )].