A particle executes SHM on a straight line path. The amplitude of oscillation is `2cm`. When the displacement of the particle from the mean position is `1cm`, the numerical value of magnitude of acceleration is equal to the mumerical value of velocity. Find the frequency of SHM (in `Hz`).

A particle executes SHM in a straight line path .The amplitude of oscillation is 2cm . When the displacement of the particle from the mean position is 1cm the numerical value of acceleration is equal to the numerical value of velocity. Then find the frequency of SHM .

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

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.

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.

A particle executes SHM on a straight line path of amplitude 2 cm. When the displacement is 1 cm, the magnitude of velocity is equal to that of acceleration. The frequency of oscillations is

A particle executes SHM on a straight line path. The amplitude of oscialltion 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. Hint : A= 3cm When x=1 cm , magnitude of velocity = magnitude of acceleration i.e., omega(A^(2) - x^(2))((1)/(2)) = omega^(2) x Find omega T = ( 2pi)(omega)

A particle executes SHM on a straight line path. The amplitude of oscialltion 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. Hint : A= 3cm When x=1 cm , magnitude of velocity = magnitude of acceleration i.e., omega(A^(2) - x^(2))((1)/(2)) = omega^(2) x Find omega T = ( 2pi)(omega)

A particle is executing SHM of amplitude 10 cm. Its time period of oscillation is pi seconds. The velocity of the particle when it is 2 cm from extreme position is

A particle executes S.H.M with amplitude of 10 cm and period of 10 s. Find the acceleration of the particle at a distance 5 cm from the equilibrium position.

A particle executes S.H.M with amplitude of 10 cm and period of 10 s. Find the velocity of the particle at a distance 5 cm from the equilibrium position.