First slide

Simple harmonic motion

Question

In a simple harmonic oscillator, at the mean position is

Easy

A

KE maximum, PE maximum
B

KE is minimum, PE is maximum
C

KE is maximum, PE is minimum
D

both KE and PE are minimum

Solution

We know $KE= \frac{1}{2} {mω}^{2} (A^{2} {-x}^{2})$
${PE=1/2mω}^{2} x^{2}$
At mean position x=0
${KE=1/2mω}^{2} A^{2}$ (maximum)
$PE=0$ (minimum).

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