The shortest distance travelled by a particle executing SHM from mean position in 2 s is equal to `(sqrt(3)//2)` times its amplitude. Determine its time period.

The shortest distance travelled by a particle executing S.H.M. from mean position in 4 seconds is eual to (1)/(sqrt(2)) times its amplitude. Find the time period. Hint : t=4 s , x = (1)/(sqrt(2)) , T=? x = A sin omega t , omega = ( 2pi)/( T ) Solve to get T.

Acceleration of a particle, executing SHM, at it’s mean position is

A particle executes SHM from extreme position and covers a distance equal to half to its amplitude in 1 s.

What is the acceleration of a particle executing S.H.M. at its mean position.?

What is the total distance travelled by a body executing SHM in a time equal to its time period, if its amplitude is A?

The displacement from mean position of a particle in SHM at 3 seconds is sqrt(3) /2 of the amplitude. Its time period will be :

When a particle executing a linear S.H.M. moves from its extreme position to the mean position, its

The time period of a particle executing S.H.M.is 12 s. The shortest distance travelledby it from mean position in 2 second is ( amplitude is a )

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

If a particle is executing SHM , with an amplitude A , the distance moved and the displacement of the body during its time period is