A particle of mass is attached to one end of a massless spring of force constant , 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 with an init

Question

A particle of mass $m$ is attached to one end of a massless 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 [2013]

Smart Achievers · Accepted Answer

(1, 4)

The particle collides elastically with rigid wall.

$∴$ $e = \frac{V}{0.5 u_{0}} = 1 \Rightarrow V = 0.5 u_{0}$

i.e., the particle rebounds with the same speed. Therefore the particle will return to its equilibrium position with speed $u_{0}$ . So, (1) is correct.

The velocity of the particle becomes $0.5 u_{0}$ after time $t$ .

Using equation $V = V_{\max} \cos ω t$

$0.5 u_{0} = u_{0} \cos ω t$

$∴$ $\frac{π}{3} = \frac{2 π}{T} \times t \Rightarrow t = \frac{T}{6}$

The time period $T = 2 π \sqrt{\frac{m}{k}}$ $∴$ $t = \frac{π}{3} \sqrt{\frac{m}{k}}$

The time taken by the particle to pass through the equilibrium for the first time $= 2 t = \frac{2 π}{3} \sqrt{\frac{m}{k}}$

So, (2) is wrong.

The time taken for the maximum compression $= t_{A B} + t_{B A} + t_{A C}$

$= \frac{π}{3} \sqrt{\frac{m}{k}} + \frac{π}{3} \sqrt{\frac{m}{k}} + \frac{π}{2} \sqrt{\frac{m}{k}}$

$= π \sqrt{\frac{m}{k}} [\frac{1}{3} + \frac{1}{3} + \frac{1}{2}]$

$= \frac{7 π}{6} \sqrt{\frac{m}{k}}$

So, (3) is wrong.

The time taken for particle to pass through the equilibrium position second time

$= 2 [\frac{π}{3} \sqrt{\frac{m}{k}}] + π \sqrt{\frac{m}{k}}$

$= π \sqrt{\frac{m}{k}} (\frac{2}{3} + 1)$

$= \frac{5 π}{3} \sqrt{\frac{m}{k}}$

So (4) is correct.

Solution

1 The speed of the particle when it returns to its equilibrium position is $u_{0}$ .

2 The time at which the particle passes through the equilibrium position for the first time is $t = π \sqrt{\frac{m}{k}} .$

3 The time at which the maximum compression of the spring occurs is $t = \frac{4 π}{3} \sqrt{\frac{m}{k}} .$

4 The time at which the particle passes through the equilibrium position for the second time is $t = \frac{5 π}{3} \sqrt{\frac{m}{k}} .$

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Solution

1 The speed of the particle when it returns to its equilibrium position is u0.

2 The time at which the particle passes through the equilibrium position for the first time is t=πmk.

3 The time at which the maximum compression of the spring occurs is t=4π3mk.

4 The time at which the particle passes through the equilibrium position for the second time is t=5π3mk.

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1 The speed of the particle when it returns to its equilibrium position is $u_{0}$ .

2 The time at which the particle passes through the equilibrium position for the first time is $t = π \sqrt{\frac{m}{k}} .$

3 The time at which the maximum compression of the spring occurs is $t = \frac{4 π}{3} \sqrt{\frac{m}{k}} .$

4 The time at which the particle passes through the equilibrium position for the second time is $t = \frac{5 π}{3} \sqrt{\frac{m}{k}} .$