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 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 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 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 linear simple harmonic motion with an amplitude of 2 cm . When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

A particle executes linear simple harmonic motion with an amplitude of 3 cm . When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then, its time period in seconds is

A particle executes simple harmonic motion with an amplitude of 5cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then its periodic time in seconds is :

A particle excutes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds 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 executies linear simple harmonic motion with an amplitude 3cm .When the particle is at 2cm from the mean position , the magnitude of its velocity is equal to that of acceleration .The its time period in seconds is

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