Question

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

1 - Find the constant of a force of the spring :

2 - Find the oscillation frequency

3 - What is the height above the initial level from which the mass m was dropped?

Answer 1

$k=m{\omega }^{2}x=b-a$

$f=kx$

Force constant : $k=\frac{2mg}{b-a}=\frac{f}{x}$

$\omega =\frac{k}{m^{\prime }}=\frac{2Mg}{b-a}\times \left(\frac{M+m}{m+M}\right)\times \frac{5}{2}\left(\frac{M+m}{m}\right)\times \frac{ab}{b-a}$

$m^{\prime }=Reducedmass=\left(\frac{mM}{m+M}\right)$

$\therefore \omega =\left(\frac{M+m}{m}\right)\frac{ab}{b-a}\Rightarrow$ Angular frequency

Height above the initial level : $h=\frac{1}{2\pi }\sqrt{km}+M$

$h=\frac{1}{2\pi }\sqrt{\frac{2mg}{(b-a)(M-m)}}\Rightarrow \frac{1}{2\pi }\sqrt{\frac{K}{M+m}}=b$