Question

One end of a spring of force constant $'k'$ is fixed to a vertical wall and the other to a block of mass $'in'$ resting on a smooth horizontal surface. There is another wall at a distance $'x_0'$ 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, $A=2 x_{0}$.
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, $x_{0}=2 x_{0} \sin \omega t \Rightarrow x_{0}=2 x_{0} \sin \frac{2 \pi}{T} t$
$\Rightarrow \quad \sin \frac{\pi}{6}=\sin \frac{2 \pi}{T} t \Rightarrow t=\frac{T}{12}$
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,
$\frac{T}{3}=\frac{2 \pi}{3} \sqrt{\frac{m}{k}}$