Two particles are executing SHM in a straight line. Amplitude A and the time period T of both the particles are equal. At time t=0, one particle is at displacement `x_(1)=+A` and the other `x_(2)=(-A/2)` and they are approaching towards each other. After what time they across each other? `T/4`

Two particles are in SHM in a straight line about same equilibrium position. Amplitude A and time period T of both the particles are equal. At time t=0 , one particle is at displacement y_(1)=+A and the other at y_(2)=-A//2 , and they are approaching towards each other. after what time they cross each other?

A particle executing SHM with time period T and amplitude A. The mean velocity of the particle averaged over quarter oscillation, is

A particle executing SHM has amplitude of 1m and time period pi sec . Velocity of particle when displacement is 0.8m is

A particle is executing simple harmonic motion with an amplitude A and time period T. The displacement of the particles after 2T period from its initial position is

A particle executes SHM with a time period T . The time period with which its potential energy changes is

particle is executing S.H.M. If its amplitude is 2 m and periodic time 2 seconds , then the maximum velocity of the particle will be

Two particles are in SHM along same line with same amplitude A and same time period T. At time t=0, particle 1 is at + A/2 and moving towards positive x-axis. At the same time particle 2 is at -A/2 and moving towards negative x-axis. Find the timewhen they will collide.

A particle is executing SHM with time period T. If time period of its total mechanical energy isT' then (T')/(T) is

A particle executing SHM of amplitude 4 cm and T=4 s .The time taken by it to move from position to half the amplitude is