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  4. 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 <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-qxrgnva9zekz.png />

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Q. 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

Question

NTA AbhyasNTA Abhyas 2020Oscillations

Solution:

Amplitude of vibration $=\frac{b + a}{2}$
Mean position $=a-\frac{b + a}{2}=\frac{a - b}{2}$
$\therefore \, k\frac{\left(a - b\right)}{2}=mg\Rightarrow k=\frac{2 m g}{\left(a - b\right)}$