Question

One end of a spring of force constant ${}^{\prime }{k}^{\prime }$ is fixed to a vertical wall and the other to a block of mass ${}^{\prime }i{n}^{\prime }$ resting on a smooth horizontal surface. There is another wall at a distance ${}^{\prime }{x}_0^{\prime }$ from the block. The spring is then compressed by $2x_0$ and released. The time taken by the block to strike the other wall is

• A. $\frac{1}{6}\pi \sqrt{\frac{m}{k}}$
• B. $\sqrt{\frac{m}{k}}$
• C. $\frac{2\pi }{3}\sqrt{\frac{m}{k}}$
• D. $\frac{\pi }{4}\sqrt{\frac{m}{k}}$

Answer 1

Answer: (C)

According to the question,

Given that the amplitude of the motion, $A=2x_0$.
We know that the time needed to cover from compressed position to mean position (normal) $=\frac{T}{4}$.
Where, $T$ is the time period of oscillation for the displacement from the mean position to $x_0$, the time taken, $t$ can be obtained from $y=A\sin \omega t$
where, $y$ is the displacement from mean position, $A$ is the amplitude.
Therefore, $x_0=2x_0\sin \omega t\Rightarrow x_0=2x_0\sin \frac{2\pi }{T}t$
$\Rightarrow \sin \frac{\pi }{6}=\sin \frac{2\pi }{T}t\Rightarrow t=\frac{T}{12}$
Therefore, the time taken to hit the wall will be,
$\frac{T}{4}+\frac{T}{12}=\frac{T}{3}$
If a mass $m$ is suspended from a spring of force constant $k$, then the time period will be,
$T=2\pi \sqrt{\frac{m}{k}}$
Hence, the time taken to hit the wall,
$\frac{T}{3}=\frac{2\pi }{3}\sqrt{\frac{m}{k}}$