Amplitude of oscillation of a particle that executes `SHM` is `2cm`. Its displacement from its mean position in a time equal to `1//6^(th)` of its time period is

A particle executes SHM of amplitude A. At what distance from the mean position is its KE equal to its PE ?

The time period of oscillation of a particle that executes SHM is 1.2s . The time starting from mean position at which its velocity will be half of its velocity at mean position is

The amplitude of oscillation of particles in SHM is sqrt(3)cm . The displacement from mean position at which its potential and kinetic energies are in the ratio 1:2 is

A particle executes SHM of amplitude A .At what distance from the mean position is its kinetic energy equal to its potential energy?

A particle executes SHM on a straight line path. The amplitude of oscillation is 3 cm. Magnitude of its acceleration is eqal to that of its velocity when its displacement from the mean position is 1 cm. Find the time period of S.H.M.

The time period of a particle executing SHM is T . After a time T//6 after it passes its mean position, then at t = 0 its :

The time period of a particle executing SHM is T . After a time T//6 after it passes its mean position, then at t = 0 its :

The time period of a particle executing SHM is T . After a time T//6 after it passes its mean position, then at t = 0 its :

A partical executes SHM on a straigh line path. The amplitude of oscillation is 2cm. When the displacement of the particle from the mean position is 1cm, the magnitude of its acceleration is equal to that of its velocity. Find the time period of SHM, also the ms. velocity and ms. acceleration of SHM.