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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

A block of mass one

\mathrm{kg}

is fastened to a spring with a spring constant

50 \mathrm{~Nm}^{-1}

. The block is pulled to a distance

x=10 \mathrm{~cm}

from its equilibrium position at

x=0

on a frictionless surface from rest at

t=0

. Write the expression for its

x(t)

and

v(t)

.

View answer and explanation

Correct option: A

0.707 \cos 7.07 t \mathrm{~ms}^{-1}

Option A (

0.707 \cos 7.07 t \text{{ ms}}^{-1}

) is correct because it represents the velocity of the block as a function of time, derived from simple harmonic motion equations.

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