A body of mass 5 kg hangs from a spring and oscillates with a time period of 2π seconds. If the ball is removed, the length of the spring will decrease by (a) g/k metres (b) k/g metres (c) 2π metres (d) g metres

Answer: (d). T = 2π√(m/k) = 2π ⇒ m/k = 1 s ². Static extension x = mg/k = g×1 = g m Removing the body removes this extension, so the answer is (d) g metres.

A body of mass 5 kg hangs from a spring and oscillates with a time… | NEET 1994 (AIPMT)

Q100NEET 1994 · AIPMTMedium

Spring Systems

A body of mass $5\text{ kg}$ hangs from a spring and oscillates with a time period of $2\pi$ seconds. If the ball is removed, the length of the spring will decrease by

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Answer: (d)

$T = 2\pi\sqrt{m/k} = 2\pi \Rightarrow \frac{m}{k} = 1$ s $^2$ .
Static extension $x = \frac{mg}{k} = g\times1 = g$ m
Removing the body removes this extension, so the answer is (d) $g$ metres.

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