Question

A particle of mass $m$ is attached to one end of a mass-less spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t$ $=$ $0$ with an initial velocity $u_0$. When the speed of the particle is $0.5$ $u_0$. It collides elastically with a rigid wall. After this collision :

• A. the speed of the particle when it returns to its equilibrium position is $u_0$
• B. the time at which the particle passes through the equilibrium position for the first time is $t=\pi \sqrt{\frac{m}{k}}$
• C. the time at which the maximum compression of the spring occurs is $t=\frac{4\pi }{3}\sqrt{\frac{m}{k}}$.
• D. the time at which the particle passes through the equilibrium position for the second time is $t=\frac{5\pi }{3}\sqrt{\frac{m}{k}}$.

Answer 1

Answer: (A), (D)

The correct options are
A the speed of the particle when it returns to its equilibrium position is $u_0$
D the time at which the particle passes through the equilibrium position for the second time is $t=\frac{5\pi }{3}\sqrt{\frac{m}{k}}$.

$v=u_0\sin \omega t$
(Suppose $t_1$ is the time of collision) $\frac{u_0}{2}=u_0\cos \omega t_1=>t_1=\frac{\pi }{3\omega }$

Now the particle returns to equilibrium position at time,
$t_2=2t_1$ i.e. $\frac{2\pi }{3\omega }$ with the same mechanical energy i.e. its speed will $u_0$.
Let $t_3$ is the time at which the particle passes through the equilibrium position for the second time.

$t_3=\frac{T}{2}+2t_1$

$=\frac{\pi }{\omega }+\frac{2\pi }{3\omega }=\frac{5\pi }{3\omega }$

$=\frac{5\pi }{3}\sqrt{\frac{m}{k}}$

Energy of the particle and spring remains conserved.