First slide

Simple harmonic motion

Question

A particle perfroms SHM with a period T and amplitude a. The mean velocity of the particle over the time interval during which it travels a distance a/2 from the extreme position is

Moderate

A

a/T
B

2a/T
C

3a/T
D

a/2T

Solution

$\begin{array}{l} x = a \cos ω t \\ a / 2 = a \cos ω t_{0} \\ or \begin{array}{ll} \cos \frac{π}{3} = \cos ω t_{0} or \frac{π}{3} = ω t_{0} ∴ t_{0} = \frac{π}{3 ω} \end{array} \\ v = \frac{d x}{d t} = - a ω \sin ω t \\ ∴ \bar{v} = \frac{\int_{0}^{π / 3 ω} v d t}{\int_{0}^{π / 3 ω} d t} \end{array}$
$But ω = \frac{2 π}{T} ∴ \bar{v} = \frac{3 a}{T}$

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