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

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" /> (A) (2 m g/b - a) (B) (m g/b - a) (C) (2 m g/a - b) (D) (m g/2 (a - b))

Tardigrade Admin · Accepted Answer

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

$\frac{2 m g}{b - a}$

$\frac{m g}{b - a}$

$\frac{2 m g}{a - b}$

$\frac{m g}{2 ( a - b )}$

Amplitude of vibration $= \frac{b + a}{2}$
Mean position $= a - \frac{b + a}{2} = \frac{a - b}{2}$
$∴ k \frac{( a - b )}{2} = m g \Rightarrow k = \frac{2 m g}{( a - b )}$

b−a2mg​

b−amg​

a−b2mg​

2(a−b)mg​

Amplitude of vibration =2b+a​ Mean position =a−2b+a​=2a−b​ ∴k2(a−b)​=mg⇒k=(a−b)2mg​

$\frac{2 m g}{b - a}$

$\frac{m g}{b - a}$

$\frac{2 m g}{a - b}$

$\frac{m g}{2 ( a - b )}$

Amplitude of vibration $= \frac{b + a}{2}$
Mean position $= a - \frac{b + a}{2} = \frac{a - b}{2}$
$∴ k \frac{( a - b )}{2} = m g \Rightarrow k = \frac{2 m g}{( a - b )}$