The time taken by a particle performing SHM to pass from point A and B where it is velocities are same is 2s. After another 2 s it returns to B. The time period oscillation is (in seconds)

The time taken by a particle performing SHM to pass from point A to B where its velocities are same is 2 s . After another 2 s it returns to B . The ratio of distance OB to its a amplitude (where O is the mean position) is

A particle performing SHM starts equilibrium position and its time period is 16 seconds. After 2 seconds its velocity is pi m//s . Amplitude of oscillation is (cos 45^(@) = 1/(sqrt(2)))

Assertion : A particle performing SHM at certain instant is having velocity v. It again acquire a velocity v for the first time after a time interval of T second. Then the time period of oscillation is T second. Reason : A particle performing SHM can have same velocity at two instants in one cycle.

A particle is performing SHM of amplitude 'A' and time period 'T'. Find the time taken by the particle to go from 0 to A//2 .

If the time taken by a wave to pass through a point is 0.002 s. What is its time period and frequency.

A pariticle is performing SHM of amplitude "A" and time period "T" . Find the time taken by the period to go from 0 to A//root 2 .

A particle performs SHM with an amplitude of 5 m. If at x = 4 m, the magnitude of velocity and accelertion are equal, then the time period of SHM (in seconds) is

The velocity of a particle is zero at time t = 2s, then

A pariticle is performing SHM of amplitude "A" and time period "T" . Find the time taken by the period to go from 0 to A//2 .

A particle performs SHM with a period T and amplitude a. The mean velocity of particle over the time interval during which it travels a//2 from the extreme position is