Question

A mass $M$ is in static equilibrium on a mass less 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 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

• A. $K=\frac{2mg}{b-a},\left(\frac{m+M}{m}\right)\frac{ab}{b-a},\frac{1}{2\pi }\sqrt{\frac{2mg}{(b-a)(M+m)}}$
• B. $K=\frac{3mg}{b-a},\left(\frac{m+M}{m}\right)\frac{ab}{b-a},\frac{1}{2\pi }\sqrt{\frac{2mg}{(b-a)(M+m)}}$
• C. $K=\frac{2mg}{b-a},\left(\frac{m+M}{2m}\right)\frac{ab}{b-a},\frac{1}{2\pi }\sqrt{\frac{2mg}{(b-a)(M+m)}}$
• D. $K=\frac{2mg}{b-a},\left(\frac{m+m}{m}\right)\frac{ab}{b-a},\frac{1}{2\pi }\sqrt{\frac{3mg}{(b-a)(M+m)}}$

Answer 1

The correct option is A $K=\frac{2mg}{b-a},\left(\frac{m+M}{m}\right)\frac{ab}{b-a},\frac{1}{2\pi }\sqrt{\frac{2mg}{(b-a)(M+m)}}$

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

$f=kx$

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: $h=\frac{1}{2\pi }\sqrt{k}m+M$

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