In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillations will be :

Simple Harmonic Motion PYQ — JEE Main 2021 · 24 Feb · Shift 1 · Q46 with Solution

View written solutionFree

Correct answer: D

Initial motion of mass $M$

A block of mass $M$ attached to a spring of constant $k$ performs SHM with amplitude $A$ .

Its angular frequency is

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

Since the second mass is attached when the block is at the equilibrium position, the displacement at that instant is

x=0.

At equilibrium, the speed is maximum, so

v_{\max}=\omega A=A\sqrt{\frac{k}{M}}.

Attaching mass $m$ gently at equilibrium

When mass $m$ is gently placed on $M$ , the two masses move together thereafter as a single mass

M+m.

Because this happens at the equilibrium position, there is no spring deformation change at that instant, so we use conservation of linear momentum during the attachment:

Mv_{\max}=(M+m)v'.

Thus,

v'=\frac{M}{M+m}v_{\max}.

Substitute $v_{\max}=A\sqrt{\frac{k}{M}}$ :

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

New SHM after attachment

Now the combined mass $M+m$ oscillates with new angular frequency

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

At the instant just after attachment, the system is still at equilibrium position $x=0$ , so its amplitude $A'$ satisfies

v'=\omega' A'.

Hence,

A'=\frac{v'}{\omega'}.

Substitute the expressions:

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

Simplify:

A'=A\frac{M}{M+m}\sqrt{\frac{k}{M}\cdot\frac{M+m}{k}}

A'=A\frac{M}{M+m}\sqrt{\frac{M+m}{M}}

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

Matching with options

Thus the new amplitude is

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

This matches Option D.

Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So, the derived answer agrees with the stored correct answer.

More from Simple Harmonic Motion

View written solutionFree

Correct answer: D

Initial motion of mass $M$

A block of mass $M$ attached to a spring of constant $k$ performs SHM with amplitude $A$ .

Its angular frequency is

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

Since the second mass is attached when the block is at the equilibrium position, the displacement at that instant is

x=0.

At equilibrium, the speed is maximum, so

v_{\max}=\omega A=A\sqrt{\frac{k}{M}}.

Attaching mass $m$ gently at equilibrium

When mass $m$ is gently placed on $M$ , the two masses move together thereafter as a single mass

M+m.

Because this happens at the equilibrium position, there is no spring deformation change at that instant, so we use conservation of linear momentum during the attachment:

Mv_{\max}=(M+m)v'.

Thus,

v'=\frac{M}{M+m}v_{\max}.

Substitute $v_{\max}=A\sqrt{\frac{k}{M}}$ :

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

New SHM after attachment

Now the combined mass $M+m$ oscillates with new angular frequency

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

At the instant just after attachment, the system is still at equilibrium position $x=0$ , so its amplitude $A'$ satisfies

v'=\omega' A'.

Hence,

A'=\frac{v'}{\omega'}.

Substitute the expressions:

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

Simplify:

A'=A\frac{M}{M+m}\sqrt{\frac{k}{M}\cdot\frac{M+m}{k}}

A'=A\frac{M}{M+m}\sqrt{\frac{M+m}{M}}

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

Matching with options

Thus the new amplitude is

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

This matches Option D.

Comparison with stored correct answer

Stored correct answer: D

Our derived answer: D

So, the derived answer agrees with the stored correct answer.

Simple Harmonic Motion question

Simple Harmonic Motion question

More from Simple Harmonic Motion

Simple Harmonic Motion question

Simple Harmonic Motion question

More from Simple Harmonic Motion