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A mass M is in static equilibrium on a m...

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?

Text Solution

Verified by Experts

The correct Answer is:
`(a)K=(2mg)/(b-a)`
`(c )((M+m)/(m))(ab)/(b-a),(1)/(2pi)sqrt((2mg)/((b-a)(M+m))`
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