A light spiral spring supports a 200 g weight at its lower end. It oscillates up and down with a period of 1 s. How much weight (in gram) must be removed from the lower end to reduce the period to 0.5 s?

Question

Dr. Abubacker · Accepted Answer

(c) Using, $T = 2\pi\sqrt{\frac{M}{k}}$, we have
$1 = 2\pi\sqrt{\frac{200}{k}}$ …(i)
and $0.5 = 2\pi\sqrt{\frac{(200 - m)}{k}}$ …(ii)
on solving, $m = 150\ g$.

Q. A light spiral spring supports a 200 g weight at its lower end. It oscillates up and down with a period of 1 s. How much weight (in gram) must be removed from the lower end to reduce the period to 0.5 s?

Explanation

Reviewed by Dr. Abubacker

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