Find the distance traveled by a particle execcuting SHM with time period T and amplitude `x_(m)` in time T/12 after starting from rest.

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

A particle executes SHM with time period T and amplitude A . The maximum possible average velocity in time T//4 is-

Amplitude of a particle executing SHM is a and its time period is T. Its maximum speed is

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

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

A particle executes SHM with a time period of 4 s . Find the time taken by the particle to go from its mean position to half of its amplitude . Assume motion of particle to start from mean position.

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.

A particle is executing SHM with time period T Starting from mean position, time taken by it to complete (5)/(8) oscillations is,

A partical excuting SHM with time period of 2 s : Find the time taken by it to move from one amplitude t half the amplitude posituion.

A particle moves along a line so that its velocity at time t is v(t)=t^(2)-t-6 (measured in meters per second) ltbr. (a). Find the displacement of the particle during the time period 1letle4 (b). Find the distance traveled dring this time period.