First slide

Simple hormonic motion

Question

A body performs SHM along the straight line segment ABCDE with C as the mid point of segment AE (A and E are the extreme position for the SHM). Its kinetic energies
at B and D arc each one fourth of its maximum value. If length of segment AE is 2R, then the distance between B and D is

Moderate

A

$\frac{\sqrt{3}}{2} R$
B

$\frac{R}{\sqrt{2}}$
C

$\sqrt{3} R$
D

$\sqrt{2} R$

Solution

When the particle crosses D, its speed is half the maximum speed
$∴ V = \frac{v_{max}}{R} \sqrt{R^{2} - x^{2}}$
$or \frac{v_{max}}{2} = \frac{v_{max}}{R} \sqrt{R^{2} - x^{2}} or x = \frac{\sqrt{3}}{2} R$
$∴ Distance BD = 2 x = \sqrt{3} R$

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