First slide

Simple hormonic motion

Question

A particle performs S.H.M. of amplitude A along a straight line. When it is at a distance $\frac{\sqrt{3}}{2}$ A from mean position, its kinetic energy gets increased by an amount $\frac{1}{2} {mω}^{2} A^{2}$ due to an impulsive force. Then its new 2 amplitude becomes

Moderate

A

$\frac{\sqrt{5}}{2} A$
B

$\frac{\sqrt{3}}{2} A$
C

A
D

$\sqrt{2} A$

Solution

Due to impulse, the total energy of the particle becomes:
$\frac{1}{2} {mω}^{2} A^{2} + \frac{1}{2} {mω}^{2} A^{2} = {mω}^{2} A^{2}$
Let ‘A’ be the new amplitude.
$∴ {mω}^{2} {(A^{'})}^{2} = {mω}^{2} A^{2}$
$\Rightarrow A^{'} = A .$

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