JEE Main 2023 · Physics · Simple Harmonic Motion · medium

A block of mass M performs SHM of amplitude A on a smooth horizontal surface, attached to a spring. At the instant it passes through the mean…

Single correct option

A block of mass

M

performs SHM of amplitude

A

on a smooth horizontal surface, attached to a spring. At the instant it passes through the mean position, a lump of putty of mass

M/3

is dropped on it from directly above and sticks. The new amplitude of oscillation is

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Answer: (D) $\dfrac{\sqrt3}{2}A$

The putty has no horizontal momentum, so horizontal momentum is conserved at the mean position:

Mv=\tfrac43Mv'

, so

v'=\tfrac34v

with

v=\omega A

. The new angular frequency is

\omega'=\sqrt{k/(4M/3)}=\omega\sqrt{3/4}

. At the mean position speed is maximum, so

A'=\dfrac{v'}{\omega'}=\dfrac{\tfrac34\omega A}{\tfrac{\sqrt3}{2}\omega}=\dfrac{\sqrt3}{2}A

. (Equivalent: energy falls to

\tfrac34

, and

A'\propto\sqrt{E'}

at fixed

k

gives the same.)

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