Question

A pendulum is undergoing S.H.M. The velocity of the bob in the mean position is v. If now its amplitude is doubled, keeping the length same, its velocity in the mean position will be(a) v/2(b) v(c) 2v(d) 4v

Answer 1

Answer:(c). 2v

Explanation:

Given that,

Velocity = v

Amplitude = a

We know that,

The velocity is

$v = \omega \sqrt{a^{2}- x^{2}}$

Now, at mean position x = 0

Therefore, the velocity is $v = \omega \sqrt{a^{2}}$

Taking square both sides

$v^{2}= \omega^{2}a^{2}$

Where, $\omega$ is constant

So, $v^{2} \propto a^{2}$

If the amplitude is double then the velocity is double because velocity is direct proportional to amplitude.

Hence, option (c) is correct.