Question

One end of a spring of force constant $k$ is fixed to a vertical wall and the other end is connected to a block of mass $m$ resting on a smooth horizontal surface. There is another wall at a distance $x_0$ from the body. The spring is now compressed by $3x_0$ and released. The time after which the block will strike the other wall is

• A. $\frac{11\pi }{9}\sqrt{\frac{m}{k}}$
• B. $\frac{2\pi }{3}\sqrt{\frac{m}{k}}$
• C. $\frac{\pi }{4}\sqrt{\frac{m}{k}}$
• D. $\frac{\pi }{6}\sqrt{\frac{m}{k}}$

Answer 1

The correct option is B $\frac{2\pi }{3}\sqrt{\frac{m}{k}}$

Amplitude of the motion,$A=2\times 0$

We know that the time needed to cover from compressed

position to mean position $=\frac{T}{4}$

For the displacement from the mean position to

$X_O$ time taken $t$ will be

$Y=Asinwt$

$\therefore X_O=2X_{O}sinwt$

but $w=\frac{2\pi }{T}$

or,$X_O=2X_{O}sin\frac{2\pi }{T}t$

or,$\frac{1}{2}=sin\frac{2\pi }{T}t$

or,$\begin{array}{c}\frac{2\pi }{T}t={\sin }^{-}\left(0.5\right)\\ \frac{2\pi }{T}t=\frac{\pi }{6}\\ t=\frac{T}{12}\end{array}$

Time taken to hit the wall$=\frac{T}{4}+\frac{T}{12}=\frac{T}{3}$

If mass M is suspended from he spring of force constant

k,$T=2\pi \sqrt{\frac{M}{R}}$

$\therefore$ Time taken to hit the wall $=\frac{T}{3}=\frac{2\pi }{3}\sqrt{\frac{M}{R}}$

Hence option $B$ is correct answer.