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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

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=\frac{A}{\sqrt{3}}

will be

View answer and explanation

Correct option: D

\frac{\pi A}{T} \sqrt{\frac{8}{3}}

Option D (

\frac{\pi A}{T} \sqrt{\frac{8}{3}}

) is correct because it is the correct application of the velocity formula in SHM at the given displacement.

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a

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. At a displacement

{y_1}

its potential energy is

{U_1}

and at a displacement

{y_2}

its potential energy is

{U_2}

. The potential energy of the particle at displacement

\left( {{y_1} + {y_2}} \right)