Question

A mass $m$ is hung on an ideal massless spring. Another equal mass is connected to the other end of the spring. The whole system is at rest. At $t=0$, $m$ is released and the system falls freely under gravity. Assume that natural length of the spring is $L_0$, its initial stretched length is $L$ and the acceleration due to gravity is $g$.
What is distance between masses as function of time ?

• A. $L_0+(L-L_0)\cos \sqrt{\frac{2k}{m}}t$
• B. $L_0+(L-L_0)\cos \sqrt{\frac{k}{m}}t$
• C. $L_0+2(L+2L_0)\cos \sqrt{\frac{2k}{m}}t$
• D. $L_0+(L-L_0)\sin \sqrt{\frac{2k}{m}}t$

Answer 1

The correct option is A $L_0+(L-L_0)\cos \sqrt{\frac{2k}{m}}t$

In CM frame both the masses execute SHM with
$\omega =\sqrt{\frac{k}{\mu }}=\sqrt{\frac{2k}{m}}$
Here reduced mass $\mu =\frac{m\left(m\right)}{2m}=\frac{m}{2}$
Initially particles are at extreme
distance and amplitude of oscillation is $L-L_0$
Hence distance between particle is
$=L_0+(L-L_0)\cos \sqrt{\frac{2k}{m}}t$