JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

One end of a compressed spring of spring constant

5 \times 10^3 \mathrm{~N m}^{-1}

is fixed to a wall and a block of mass

0.5 \mathrm{~kg}

is kept in contact with the other end of the spring. If the spring is suddenly released, the block moves with a momentum

10 \mathrm{~Ns}

. Then the compression of the spring is

View answer and explanation

Correct option: B

20 \mathrm{~cm}

Option B (

20 \mathrm{~cm}

) is correct because it is calculated by equating the kinetic energy of the block to the potential energy stored in the compressed spring and solving for the compression.

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{y_1}

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{U_1}

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{y_2}

its potential energy is

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