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JEE Main 202124 FebruaryMorning Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Morning Shift
Subject: Physics
Chapter: Oscillations

JEE Main 2021 · 24 February · Morning Shift · Physics · Oscillations

In the given figure, a mass M is attached to a horizontal spring...

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 oscillation will be:

In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The

Options

A $A\sqrt{\dfrac{M + m}{M}}$
B $A\sqrt{\dfrac{M}{M - m}}$
C $A\sqrt{\dfrac{M - m}{M}}$
D $A\sqrt{\dfrac{M}{M + m}}$

Correct answer

D. $A\sqrt{\dfrac{M}{M + m}}$

Step-by-step solution

In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The

Momentum of system remains conserved.

${p}_{i} = {p}_{f}$

$MA\omega = \left(m + M\right){A}^{'}\omega$

$MA\sqrt{\dfrac{k}{M}} = \left(m + M\right){A}^{'}\sqrt{\dfrac{k}{m + M}}$

${A}^{'} = A\sqrt{\dfrac{M}{M + m}}$

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