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JEE PHYSICS MCQ: OSCILLATIONS

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Application +4 | -1

A spring is stretched by

0.40 \mathrm{~m}

when a mass of

0.6 \mathrm{~kg}

is suspended from it. The period of oscillations of the spring loaded by

255 \mathrm{~gm}

and put to oscillations is close to__

\left(\mathrm{g}=10 \mathrm{~m}.\mathrm{s}^{-2}\right)

View answer and explanation

Correct option: C

0.82 \mathrm{~s}

Option C (0.82 s) is correct because the period of oscillation is calculated using the formula T = 2*pi*sqrt(m/k), where k is the spring constant.

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