First slide

Simple hormonic motion

Question

A simple pendulum performs simple harmonic motion about X=0 with an amplitude A and time period T. The speed of the pendulum at
X = A /2 will be

Moderate

A

$πA \sqrt{3} / T$
B

$πA / T$
C

$πA \sqrt{3} / 2 T$
D

$3 π^{2} A / T$

Solution

$v = ω \sqrt{(A^{2} - X^{2})} = ω \sqrt{(A^{2} - A^{2} / 4)} = \sqrt{3} ωA / 2 = \sqrt{3} (\frac{2 π}{T}) A / 2 = πA \sqrt{3} / T$

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