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JEE PHYSICS MCQ: OSCILLATIONS

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A particle oscillates simple harmonically with a period of

16 \mathrm{~s}

. Two second after crossing the equilibrium position its velocity becomes

1 \mathrm{~m} / \mathrm{s}

. The amplitude is

View answer and explanation

Correct option: B

\frac{8 \sqrt{2}}{\pi} \mathrm{m}

Option B (

\frac{8 \sqrt{2}}{\pi} \mathrm{m}

) is correct because the amplitude is

\frac{8 \sqrt{2}}{\pi} \mathrm{m}

, calculated using the given period and velocity at a specific time.

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x

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y

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a

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SHM

. 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)