First slide

Simple hormonic motion

Question

Two simple harmonic oscillators with amplitudes in the ratio 1 : 2 are having the same total energy. The ratio of their frequencies is

Easy

A

1: 4
B

1: 2
C

2 : 1
D

4 : 1

Solution

$\large T.E\, = \,\frac{1}{2}m{\omega ^2}{A^2};\,T.{E_1}\, = \,T.{E_2}$
$\large \frac{1}{2}m\omega _1^2A_1^2\, = \,\frac{1}{2}m\omega _2^2A_2^2$
$\large \omega _1^2A_1^2\, = \,\omega _2^2A_2^2\,;\,\,\frac{{{w_1}}}{{{w_2}}}\, = \,\frac{{{A_2}}}{{{A_1}}}\, = \,\frac{2}{1}$

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