A mass `M` is in static equilibrium on a massless vertical spring as shown in the figure. `A` ball of mass `m` dropped from certain height sticks to the mass `M` after colliding with it.
The oscillations they perform reach to height'a' above the original level of scales `&` depth `'b'` below it.

(a) Find the constant of force of the spring.,
(b) Find the oscillation frequency.
(c ) What is the height above the initial level from which the mass `m` was dropped?

Answer

A, B, C

Explanation

`Ans : (a) K=(2mg)/(b-a)`
`(b) (1)/(2pi)sqrt((k)/(M+m))`
`(c)((M+m)/(b-a))(ab)/(m)`

Amplitude `=b-(mg)/(K)=a+(mg)/(K)(` by diagram `)`
`rArr K=(2mg)/(b-a)`
`(b) f=(1)/(T)=(1)/(2pi)sqrt((M)/(K+m))`
`(c )` If the ball was fropped from `'h'`.
`rArr V _("before collision")=sqrt(2gh)`
conserving momentum `:`
`msqrt(2gh)=(m+M)V'` Where `V'`
`=omegasqrt(A^(2)-x^(2))=sqrt((K)/(M+m))sqrt((b-(mg)/(K))^(2)-((mg)/(K))^(2))`
Squaring both sides and putting
`K=(2mg)/(b-a),`get `h=((M+m)/(b-a))(ab)/(m)`.