Question

A block of mass $m$ compresses a spring of stiffness constant $k$ and natural length $l$, through a distance $\frac{l}{2}$ as shown in the figure. If the block is not fixed with the spring, the period of motion of the block is
[Assume collision to be elastic with the wall and horizontal surface to be smooth]

• A. $2\pi \sqrt{\frac{m}{K}}$
• B. $(\pi +4)\sqrt{\frac{m}{K}}$
• C. $(1+\pi )\sqrt{\frac{m}{k}}$
• D. None of these

Answer 1

The correct option is B $(\pi +4)\sqrt{\frac{m}{K}}$

We know that, the time period of oscillation $T=2\pi \sqrt{\frac{m}{k}}$
The time period for which the block is in contact with the spring $t=\pi \sqrt{\frac{m}{k}}........\left(1\right)$
After which,it leaves the spring with a speed $v=\omega A$
From the data given in the question,
$v=\left(\sqrt{\frac{k}{m}}\right)\left(\frac{l}{2}\right).......\left(2\right)$
Since the floor is frictionless, it moves with constant velocity $v$ for a distance $D=l+l=2l$
Corresponding time taken during the motion $t_2=2l/v$
Using $\left(2\right)$ in the above equation we get,
$t_2=\frac{2l}{\frac{l}{2}\sqrt{\frac{k}{m}}}=4\sqrt{\frac{m}{k}}$
$\therefore$ The Total time period of motion $t=t_1+t_2$
$=\pi \sqrt{\frac{m}{k}}+4\sqrt{\frac{m}{k}}$
$=\sqrt{\frac{m}{k}}[\pi +4]$
Hence, option (b) is the correct answer.