Gujarati
13.Oscillations
normal

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 collies 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 througout the equilibrium position for the second time is $t=\frac{5 \pi}{3} \sqrt{\frac{ m }{ k }}$.

A

$(B,D)$

B

$(B,C)$

C

$(A,C)$

D

$(A,D)$

(IIT-2013)

Solution

Displacement $x = A \sin \omega t$

Velocity $v = A \omega \cos \omega t =\frac{\omega A }{2}$

At the time of collision

$\cos \omega t=\frac{1}{2} $

$\omega t=\frac{\pi}{3} \Rightarrow t=\frac{2 \pi}{3}=\frac{\pi}{3} \sqrt{\frac{m}{k}}$

$Image$

for $(C)$ $ \quad time$$ =\frac{2 \pi}{3} \sqrt{\frac{m}{k}}+\frac{\pi}{2} \sqrt{\frac{m}{k}} $

$ =\frac{5 \pi}{6} \sqrt{\frac{m}{k}}$(So it is incorrect) 

for $(D)$ $\quad time  \quad=\frac{2 \pi}{3} \sqrt{\frac{m}{k}}+\pi \sqrt{\frac{m}{k}} $

$ =\frac{5 \pi}{3} \sqrt{\frac{m}{k}} \text { (So it is correct). }$

Standard 11
Physics

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