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.

IIT-JEE 2011 – Paragraph Type Multiple Choice (Single Answer) Questions Involving Phase Space Diagrams

“The earth provides enough to satisfy every man’s needs, but not every man’s greed.”

– Mahatma Gandhi

The practice of asking two or three multiple choice questions (often single answer type) based on a given paragraph, is resorted to in many entrance examinations. The following three questions[(1), (2) and (3)] which appeared in IIT-JEE 2011 question paper are relatively simple:

Paragraph for questions (1), (2) and (3)

Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one–dimension. For such system, phase space is a plane is which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x (t) vs p(t) curve in this plane. The arrow on the curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.

(1) The phase space diagram for a ball thrown vertically up from the ground is

The arrow on the diagram shows the flow of time. Initially the ball is on the ground and its velocity and momentum have maximum positive values (since it moves upwards). At the highest point of its trajectory the displacement of the ball has maximum positive value but its momentum is zero. Finally when the ball just hits the ground, its displacement is zero but its momentum has maximum negative value(since it is moving downwards).

The correct option is obviously (D).

(2) The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and E₁ and E₂ are the total mechanical energies respectively. Then

(A) E₁ = √2 E₂

(B) E₁ = 2 E₂

(C) E₁ = 4 E₂

(D) E₁ = 16 E₂

When the total energy of the simple harmonic motion (SHM) is E₁, the amplitude is 2a and when the total energy is E₂, the amplitude is a, as is evident from the phase space diagrams. The total energy E of a particle of mass m in simple harmonic motion of amplitude A and angular frequency ω is given by

E = ½ mω²A²

So when the amplitude is doubled, the total energy is quadrupled.

Thus E₁ = 4 E₂ as given in option (C).

[The phase space diagram given in the above question represents an undamped simple harmonic motion since respective amplitudes appropriate to the initial conditions remain constant. If the simple harmonic motion is a damped one, the curve will be a spiral, proceeding inwards].

(3) Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is

The correct option is (B) on account of the following reasons:

(i) The amplitude of oscillations must decrease with time [which is not the case in diagrams (C) and (D)] as the viscous forces in water damps the motion of the system.

(ii) When the displacement is maximum positive value (corresponding to maximum positive position co-ordinate), the momentum is zero and is going to increase in magnitude in the negative direction [unlike in the case of diagram (A)].

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.

AIPMT 2009 Multiple Choice Questions on Work, Energy & Power

How strange is the lot of we mortals! Each of us is here for a brief sojourn; for what purpose we know not, though sometimes sense it. But we know from daily life that we exist for other people first of all for whose smiles and well-being our own happiness depends.

–Albert Einstein

Three questions from the section ‘work, energy & power’ were included in the AIPMT 2009 question paper. Here are those questions with solution:

(1) A block of mass M is attached to the lower end of a vertical spring. The spring is hung from the ceiling and has force constant value k. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be:

(1) 2 Mg/k

(2) 4 Mg/k

(3) Mg/2k

(4) Mg/k

If x is the maximum extension produced, we have

Mgx = ½ kx², on equating the decrease in the gravitational potential energy of the mass M to the increase in the elastic potential energy of the spring.

Therefore, x = 2Mg/k

(2) A body of mass 1 kg is thrown upwards with a velocity 20 m/s. It momentarily comes to rest after attaining a height of 18 m. How much energy is lost due to air friction? (g = 10 ms^–2)

(1) 30 J

(2) 40 J

(3) 10 J

(4) 20 J

The initial kinetic energy of the 1 kg mass is given by

½ mv² = ½ ×1×20² = 200 J.

The gravitational potential energy of the 1 kg mass at its maximum height is given by

Mgh = 1×10×18 = 180 J.

The energy lost due to air friction is therefore equal to 200 – 180 = 20 J.

(3) An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

(1) mv²

(2) ½ mv²

(3) ½ m²v²

(4) ½ mv³

Mas of water flowing out per second through the hose is mv. Therefore, kinetic energy imparted per second to the water is ½ ×(mv)×v² = ½ mv³ [Option (3)].

You will find useful posts in this section here.

IIT-JEE 2008 Straight Objective Type (Single Answer Multiple Choice) Questions on Conservation of Energy

The following questions involving the law of conservation of energy are simple even though they may appear to be not so at the first reading:

(1) A block (B) is attached to two unstretched springs S₁ and S₂ with spring constants k and 4 k respectively (see fig.1). The other ends are attached to identical supports M₁ and M₂ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (fig. 2) and released. The block returns and moves a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equilibrium position of the block B. The ratio y/x is

(A) 4

(B) 2

(C) ½

(d) ¼

On displacing the block B towards wall 1, spring S₁ gets compressed through x and acquires potential energy ½ kx². When the spring S₁ springs back to its original unstretched condition, it pushes the block B towards wall 2 and compresses the spring S₂ through y. In this compressed condition of spring S₂ the entire kinetic energy of the block B is transferred to spring S₂. Since the spring S₁ is free to move with the block B, it is unstretched and hence we have

½ kx² = ½ (4k) y²

This gives y/x = ½

(2) A bob of mass m is suspended by a massless string of lengh L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies

(A) θ = π/4

(B) π/4 < θ < π/2

(C) π/2 < θ < 3π/4

(D) 3π/4 < θ < π

The sum of the kinetic energy and potential energy of the of the bob in the displaced position must be equal to the kinetic energy at the position A. Therefore we have

½ M (V/2)² + MgL(1 – cos θ) = ½ MV²

The critical velocity (for just tracing the vertical circle) V = √(5gR) = √(5gL)

Substituting this value we obtain

gL(1 – cos θ) = 15gL/8 so that cos θ = – 7/8

Therefore 3π/4 < θ < π.