A point particle of mass 0.1 kg is executing SHM of amplitude 0.1$m$. When the particle passes through the mean position, its KE is $8 \times {10^{ - 3}}{\rm{J}}$. The equation of motion of this particle, if its initial phase of oscillation is ${45^0}$ is

$y = 0.1\sin \left( {\frac{r}{4} + \frac{\pi }{4}} \right)$

$y = 0.1\sin \left( {\frac{t}{2} + \frac{\pi }{4}} \right)$

JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

A point particle of mass 0.1 kg is executing SHM of amplitude 0.1

m

. When the particle passes through the mean position, its KE is

8 \times {10^{ - 3}}\mathrm{J}

. The equation of motion of this particle, if its initial phase of oscillation is

{45^0}

View answer and explanation

Correct option: D

y = 0.1\sin \left( {4t + \frac{\pi }{4}} \right)

Option D (

y = 0.1\sin \left( {4t + \frac{\pi }{4}} \right)

) is correct because the equation represents SHM with the calculated angular frequency and given initial phase.

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x

y

a

SHM

{y_1}

{U_1}

{y_2}

{U_2}

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