An ideal spring supports a disc of mass M . A body of mass m is released from a certain height from where it falls to hit M . The two masses stick together at the moment they touch and move together f

Question

An ideal spring supports a disc of mass $M$ . A body of mass $m$ is released from a certain height from where it falls to hit $M$ . The two masses stick together at the moment they touch and move together from then on. The oscillations reach to a height $a$ above the original level of the disc and depth $b$ below it. The constant of the force of the spring is

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Answer:

C

Solution:

Amplitude of vibration $= \frac{b + a}{2}$

Mean position $= a - \frac{b + a}{2} = \frac{a - b}{2}$

$∴ k \frac{(a - b)}{2} = m g \Rightarrow k = \frac{2 m g}{(a - b)}$

TestHub · Accepted Answer

Amplitude of vibration = b + a 2 Mean position = a - b + a 2 = a - b 2 ∴ k a - b 2 = m g ⇒ k = 2 m g a - b

Physics - SHM Question with Solution | TestHub

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