Acceleration of a particle in SHM at displacement `x=10 cm` (from the mean position is `a =-2.5 cm//s^(2)`). Find time period of oscillations.

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 acceleration of a particle performing SHM is 12cm//s^2 at a distance of 3 cm from the mean position. Calculate its time-period.

Maximum acceleration of a particle in SHM is 16 cm//s^(2) and maximum uelocity is 8 cm//s .Find time period and amplitude of oscillations.

Maximum acceleration of a particle in SHM is 16 cm//s^(2) and maximum uelocity is 8 cm//s .Find time period and amplitude of oscillations.

The acceleration of a particle performing SHM is 12 cm //s^(2) at a distance of 3 cm from the mean position . Calculate its time - period . Hint : a = omega^(2) x a = 12 cm//s^(2) , x=3cm Find omega T= (pi)/(omega)

The acceleration of a particle performing SHM is 12 cm //s^(2) at a distance of 3 cm from the mean position . Calculate its time - period . Hint : a = omega^(2) x a = 12 cm//s^(2) , x=3cm Find omega T= (pi)/(omega)

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

The acceleration of a particle performing S.H.M. is 12 cm/ s^(2) at a distance of 3 cm from the mean position. Its time period 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)