A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x0. Consider two cases: (i) when the block is at x0; and (ii) when the block is at x = x0 + A. In both cases, a particle with mass m( < M) is softly placed on the block after which they stick on each other. Which of the following statement(s) is(are) true about the motion after the mass m is placed on the mass M?

The amplitude of oscillation in the first case changes by a factor of \sqrt M \over m + M , whereas in the second case it remains unchanged., The final time period of oscillation in both the cases is same, The instantaneous speed at x0 of the combined masses decreases in both the cases

The amplitude of oscillation in the first case changes by a factor of \sqrt M \over m + M , whereas in the second case it remains unchanged.

The final time period of oscillation in both the cases is same

The total energy decreases in both the cases

The instantaneous speed at x0 of the combined masses decreases in both the cases

Simple Harmonic Motion PYQ — JEE Advanced 2016 · Shift 2 · Q37 with Solution

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Correct answer: A, B, D

Initial SHM of block $M$

The block of mass $M$ attached to spring constant $k$ executes SHM about equilibrium position $x_0$ with amplitude $A$ .

Its angular frequency is

\omega = \sqrt{\frac{k}{M}}.

After a mass $m$ is softly placed and sticks to it, the new mass becomes

M' = M+m,

and the new angular frequency is

\omega' = \sqrt{\frac{k}{M+m}}.

Hence the new time period is

T' = 2\pi\sqrt{\frac{M+m}{k}}.

This depends only on the final mass and spring constant, not on where the mass is placed.

So in both cases, the final time period is the same.

Therefore, B is true.

Case (i): mass $m$ placed when block is at $x=x_0$

At equilibrium position, displacement from mean is zero, so the speed of mass $M$ is maximum:

v_0 = \omega A = A\sqrt{\frac{k}{M}}.

The mass $m$ is softly placed, so initially it has zero horizontal velocity. Since sticking happens instantaneously, horizontal momentum is conserved during the placing process:

Mv_0 = (M+m)v'.

Thus,

v' = \frac{M}{M+m}v_0.

Just after sticking, the system is at equilibrium position $x_0$ , so displacement is zero and all energy is kinetic. For the new SHM, maximum speed is related to amplitude $A_1$ by

v' = \omega' A_1.

Therefore,

A_1 = \frac{v'}{\omega'} = \frac{M}{M+m} \cdot \frac{v_0}{\omega'}.

But $v_0=\omega A$ , so

A_1 = \frac{M}{M+m}\cdot \frac{\omega}{\omega'} A.

Now,

\frac{\omega}{\omega'} = \sqrt{\frac{k/M}{k/(M+m)}} = \sqrt{\frac{M+m}{M}}.

Hence,

A_1 = \frac{M}{M+m}\sqrt{\frac{M+m}{M}}A = A\sqrt{\frac{M}{M+m}}.

So in case (i), amplitude changes by factor

\sqrt{\frac{M}{M+m}}.

Case (ii): mass $m$ placed when block is at $x=x_0+A$

At extreme position, the instantaneous speed of block is zero.

So just before and just after placing the mass,

v=0.

Also the displacement from equilibrium is still

x-x_0 = A.

Now the new system of mass $M+m$ starts from rest at displacement $A$ . That is exactly an extreme position for the new SHM as well. Therefore the new amplitude remains

A_2 = A.

So in case (ii), amplitude remains unchanged.

Thus A is true.

Check total energy in both cases

Case (i)

Initial total energy:

E_i = \frac12 kA^2.

At $x_0$ , all of it is kinetic.

After sticking, the new amplitude is

A_1 = A\sqrt{\frac{M}{M+m}}.

So final energy is

E_f^{(i)} = \frac12 kA_1^2 = \frac12 kA^2\frac{M}{M+m}.

Since

\frac{M}{M+m}<1,

energy decreases.

Case (ii)

Initially at extreme position, energy is

E_i = \frac12 kA^2.

After placing mass, amplitude remains $A$ , so final energy is

E_f^{(ii)} = \frac12 kA^2.

Thus energy remains unchanged.

So the statement "total energy decreases in both cases" is false.

Therefore, C is false.

Speed at $x_0$ after placing mass

We compare the speed of the combined mass when it later passes through $x_0$ .

Case (i)

Just after sticking at $x_0$ , its speed is already

v_1 = \frac{M}{M+m}v_0,

which is less than the original $v_0$ . Thus speed at $x_0$ decreases.

Case (ii)

Amplitude remains $A$ , but new angular frequency is smaller:

\omega' = \sqrt{\frac{k}{M+m}} < \sqrt{\frac{k}{M}}=\omega.

So speed at mean position becomes

v_2 = \omega' A = A\sqrt{\frac{k}{M+m}}.

Originally it was

v_0 = A\sqrt{\frac{k}{M}}.

Since $M+m > M$ ,

v_2 < v_0.

So speed at $x_0$ decreases in this case also.

Therefore, D is true.

Final evaluation of options

A: True
B: True
C: False
D: True

So the correct options are

\boxed{A,\ B,\ D}

Comparison with stored correct answer

Stored correct answer: A, B, D

This matches our derived answer exactly.

More from Simple Harmonic Motion

View written solutionFree

Correct answer: A, B, D

Initial SHM of block $M$

The block of mass $M$ attached to spring constant $k$ executes SHM about equilibrium position $x_0$ with amplitude $A$ .

Its angular frequency is

\omega = \sqrt{\frac{k}{M}}.

After a mass $m$ is softly placed and sticks to it, the new mass becomes

M' = M+m,

and the new angular frequency is

\omega' = \sqrt{\frac{k}{M+m}}.

Hence the new time period is

T' = 2\pi\sqrt{\frac{M+m}{k}}.

This depends only on the final mass and spring constant, not on where the mass is placed.

So in both cases, the final time period is the same.

Therefore, B is true.

Case (i): mass $m$ placed when block is at $x=x_0$

At equilibrium position, displacement from mean is zero, so the speed of mass $M$ is maximum:

v_0 = \omega A = A\sqrt{\frac{k}{M}}.

The mass $m$ is softly placed, so initially it has zero horizontal velocity. Since sticking happens instantaneously, horizontal momentum is conserved during the placing process:

Mv_0 = (M+m)v'.

Thus,

v' = \frac{M}{M+m}v_0.

Just after sticking, the system is at equilibrium position $x_0$ , so displacement is zero and all energy is kinetic. For the new SHM, maximum speed is related to amplitude $A_1$ by

v' = \omega' A_1.

Therefore,

A_1 = \frac{v'}{\omega'} = \frac{M}{M+m} \cdot \frac{v_0}{\omega'}.

But $v_0=\omega A$ , so

A_1 = \frac{M}{M+m}\cdot \frac{\omega}{\omega'} A.

Now,

\frac{\omega}{\omega'} = \sqrt{\frac{k/M}{k/(M+m)}} = \sqrt{\frac{M+m}{M}}.

Hence,

A_1 = \frac{M}{M+m}\sqrt{\frac{M+m}{M}}A = A\sqrt{\frac{M}{M+m}}.

So in case (i), amplitude changes by factor

\sqrt{\frac{M}{M+m}}.

Case (ii): mass $m$ placed when block is at $x=x_0+A$

At extreme position, the instantaneous speed of block is zero.

So just before and just after placing the mass,

v=0.

Also the displacement from equilibrium is still

x-x_0 = A.

Now the new system of mass $M+m$ starts from rest at displacement $A$ . That is exactly an extreme position for the new SHM as well. Therefore the new amplitude remains

A_2 = A.

So in case (ii), amplitude remains unchanged.

Thus A is true.

Check total energy in both cases

Case (i)

Initial total energy:

E_i = \frac12 kA^2.

At $x_0$ , all of it is kinetic.

After sticking, the new amplitude is

A_1 = A\sqrt{\frac{M}{M+m}}.

So final energy is

E_f^{(i)} = \frac12 kA_1^2 = \frac12 kA^2\frac{M}{M+m}.

Since

\frac{M}{M+m}<1,

energy decreases.

Case (ii)

Initially at extreme position, energy is

E_i = \frac12 kA^2.

After placing mass, amplitude remains $A$ , so final energy is

E_f^{(ii)} = \frac12 kA^2.

Thus energy remains unchanged.

So the statement "total energy decreases in both cases" is false.

Therefore, C is false.

Speed at $x_0$ after placing mass

We compare the speed of the combined mass when it later passes through $x_0$ .

Case (i)

Just after sticking at $x_0$ , its speed is already

v_1 = \frac{M}{M+m}v_0,

which is less than the original $v_0$ . Thus speed at $x_0$ decreases.

Case (ii)

Amplitude remains $A$ , but new angular frequency is smaller:

\omega' = \sqrt{\frac{k}{M+m}} < \sqrt{\frac{k}{M}}=\omega.

So speed at mean position becomes

v_2 = \omega' A = A\sqrt{\frac{k}{M+m}}.

Originally it was

v_0 = A\sqrt{\frac{k}{M}}.

Since $M+m > M$ ,

v_2 < v_0.

So speed at $x_0$ decreases in this case also.

Therefore, D is true.

Final evaluation of options

A: True
B: True
C: False
D: True

So the correct options are

\boxed{A,\ B,\ D}

Comparison with stored correct answer

Stored correct answer: A, B, D

This matches our derived answer exactly.

Simple Harmonic Motion question

Simple Harmonic Motion question

Case (i)

Case (ii)

Case (i)

Case (ii)

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