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 u₀ . When the speed of the particle is 0.5 u_0, it collides elastically with a rigid wall. After this collision,

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

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

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

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

answer is A, D.

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Detailed Solution

$v = {ul}_{0} \cos ω t$ (suppose t₀₁ is the time of collision)

$\frac{u_{0}}{2} = u_{0} \cos {ωt}_{1} \Rightarrow t_{1} = \frac{π}{3 ω}$ Now the particle returns to equilibrium position at time $t_{2} = 2 t_{1}, i.e., \frac{2 π}{3 ω}$ with the same mechanical energy, i.e., its speed will be u₀ . Let t be the time at which the particle passes through the equilibrium position for the second time.
Energy of particle and spring remains conserved.