First slide

Simple hormonic motion

Question

A body is executing SHM under the action of a force whose maximum magnitude is 50 N. The magnitude of force acting on the particle at the time when its energy is half kinetic and half-potential is (Assume potential energy to be zero at mean position).

Moderate

A

12.5 $\sqrt{2} N$
B

12.5 N
C

25 N
D

$25 \sqrt{2} N$

Solution

$F_{\max} = {mω}^{2} A = 50 N$
Given $\frac{1}{2} {kx}^{2} = \frac{1}{2} [\frac{1}{2} {kA}^{2}] \Rightarrow x = \frac{A}{\sqrt{2}}$
$F = {mω}^{2} x = \frac{{mω}^{2} A}{\sqrt{2}} = \frac{50}{\sqrt{2}} = 25 \sqrt{2} N$

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