JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Elasticity & SHM

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 $u_{0}$ .✓ Correct
(B)the time at which the particle passes through the equilibrium position for the first time is $t = π \frac{m}{k}$ .
(C)the time at which the maximum compression of the spring occurs is $t = \frac{4 π}{3} \frac{m}{k}$ .
(D)the time at which the particle passes through the equilibrium position for the second time is $t = \frac{5 π}{3} \frac{m}{k}$ .✓ Correct

Answer: (A), (D)

Solution

Initial energy:

\frac{1}{2} m u_{0}^{2} = \frac{1}{2} k A^{2}

, so amplitude

A = u_{0} \frac{m}{k}

.

At collision, speed is

0.5 u_{0}

. Energy at collision:

\frac{1}{2} m (0.5 u_{0})^{2} = \frac{3}{8} m u_{0}^{2}

.

This equals

\frac{1}{2} k x^{2}

, giving

x = \frac{3}{2} u_{0} \frac{m}{k} = \frac{3}{2} A

.

Time to reach this position:

t_{1} = \frac{π}{3 ω} = \frac{π}{3} \frac{m}{k}

.

After elastic collision, particle returns with speed

0.5 u_{0}

. By energy conservation, when it reaches equilibrium, speed is

u_{0}

. ✓ (A)

Time to first reach equilibrium after collision:

t = \frac{π}{3} \frac{m}{k} + \frac{π}{3} \frac{m}{k} = \frac{2 π}{3} \frac{m}{k}

.

Second time at equilibrium:

t = \frac{2 π}{3} \frac{m}{k} + \frac{T}{2} = \frac{2 π}{3} \frac{m}{k} + π \frac{m}{k} = \frac{5 π}{3} \frac{m}{k}

✓ (D)

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JEE Advanced · 2013 · 02 Jun 2013 · Paper 2 · Elasticity & SHM

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