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

Engineering

Physics

Basics of Simple Harmonic Motion

Collision

Question

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

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 $t = π \sqrt{\frac{m}{k}}$

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

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\frac{ωA}{2} = ω \sqrt{A^{2} - x^{2}}$

$x = \frac{\sqrt{3} A}{2}$

$ωt = \frac{2 π}{3}$

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

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

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