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  • A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at \text{x}=\frac{\text{a}}{2} will be:

Option: 1

\frac{\pi\text{a}}{\text{T}}


Option: 2

\frac{3\pi^{2}\text{a}}{\text{T}}


Option: 3

\frac{\pi\text{a}\sqrt{3}}{\text{T}}


Option: 4

\frac{\pi\text{a}\sqrt{3}}{2\text{T}}


Answers (1)

best_answer

The velocity of a particle executing S.H.M is given by 

v = \omega\sqrt{a^{2}-x^{2}} = \frac{2\pi}{T} \sqrt{A^{2} - \frac{A^{2}}{4}} = \frac{2\pi}{T} \sqrt{\frac{3A^{2}}{4}} = \frac{\pi A\sqrt{3}}{T}

 

Posted by

Shailly goel

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