First slide

Simple harmonic motion

Question

A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic

Easy

A

1 cm
B

$\sqrt{2} cm$
C

3 cm
D

$2 \sqrt{2} cm$

Solution

Let x be the point where K.E. = P.E.
Hence $\frac{1}{2} {mω}^{2} (a^{2} - x^{2}) = \frac{1}{2} {mω}^{2} x^{2}$
$\Rightarrow 2 x^{2} = a^{2}$ $\Rightarrow x = \frac{a}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2} cm$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App