Test Index

Impulse & Momentum

Section: Physics

Question : 8 of 16

Marks: +1, -0

u_0

0.5u_0

the speed of the particle when it returns to its equilibrium position is $u_0$
the time at which the particle passes through the equlibrium position for the first time is $t = \Pi \sqrt{\frac{m}{k}}$
the time at which the maximum compression of the spring occurs is $t = \frac{4\Pi}{3} \sqrt{\frac{m}{k}}$
the time at which the particle passes throughout the equilibrium position for the second time is $t = \frac{5\Pi}{3} \sqrt{\frac{m}{k}}$

Solution:

v = \mu_0

sin ωt (suppose

t_1

is the time of collision)

\frac{\mu_0}{2} = \mu_0 \cos \omega t_1 \Rightarrow t_1 = \frac{\pi}{3\omega}

Now the particle returns to equilibrium position at time

t_2 = 2t_1 \text{ i.e } \frac{2\pi}{3\omega}

with the same mechanical energy

i.e. its speed will

\mu_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 particle and spring remains conserved.

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