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

  • the speed of the particle when it returns to its equilibrium position is u0
  • the time at which the particle passes through the equilibrium position for the first time is t = π√(m/k)
  • the time at which the maximum compression of the spring occurs is t = 4π/(3√(k/m))
  • the time at which the particle passes through the equilibrium position for the second time is t = 5π/(3√(k/m))

the time at which the particle passes through the equilibrium position for the first time ist=π√(m/k)

the time at which the maximum compression of the spring occurs ist=4π/(3√(k/m)).

the speed of the particle when it returns to its equilibrium position isu0

the time at which the particle passes through the equilibrium position for the second time ist=5π/(3√(k/m)).

Solution:

v = u0sinωt (Suppose t1 is the time of collision)
u0²/2 = u0²cos²ωt1 => t1 = π/(3ω)
Now the particle returns to equilibrium position at time, t2 = 2t1 i.e. 2π/(3ω) with the same mechanical energy i.e. its speed will be u0.
Let t3 is the time at which the particle passes through the equilibrium position for the second time.
t3 = T/2 + 2t1 = π/ω + 2π/(3ω) = 5π/(3ω) = 5π/(3√(k/m))
Energy of the particle and spring remains conserved.