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JEE PHYSICS MCQ: OSCILLATIONS

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The instantaneous displacement of a simple pendulum oscillator is given by

x=A \cos \left(\omega t+\frac{\pi}{4}\right)

. Its speed will be maximum at time :-

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Correct option: A

\frac{\pi}{4 \omega}

Option A (

\frac{\pi}{4 \omega}

) is correct because it represents the time at which the velocity function reaches its maximum value.

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Assertion: A particle executing simple harmonic motion comes to rest at the extreme positions.

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a

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{y_1}

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{y_2}

its potential energy is

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. The potential energy of the particle at displacement

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