JEE Advanced2013PhysicsOscillationsActual
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 :
Options
- AThe speed of the particle when it returns to its equilibrium position is u 0
- BThe time at which the particle passes through the equilibrium position for the first time is t = π m k
- CThe time at which the maximum compression of the spring occurs is t = 4 π 3 m k
- DThe time at which the particle passes through the equilibrium position for the second time is t = 5 π 3 m k
Correct answer
A. The speed of the particle when it returns to its equilibrium position is u 0
Step-by-step solution
Let the equation of S H M be, x = A sin ω t Here, x is the position of the particle at any time t A is the amplitude of the S H M ω is the Angular frequency of S H M So, equation for the velocity will be, v = A ω cos ω t ⇒ v = u 0 cos ω t     ( since   velocity   at   t = 0   is   u 0 ,   so   ωA = u 0 ) Now, during collision speed of particle was ( u 1 ) = 0 . 5 u 0 and time be t 1 So, Substituting the values in equation of velocity w