A particle executes simple harmonic motion of period `16 s`. Two seconds later after it passes through the center of oscillation its velocity is found to be `2 m//s`. Find the amplitude.

A particle executes SHM of period 12s. Two sec after it passes through the centre of oscillation, the velocity is found to be 3.142 cm s^(-1) find the amplitude and the length of the path.

A particle executes SHM of period 12s. Two sec after it passes through the centre of oscillation, the velocity is found to be 3.142 cm s^(-1) fin dht amplitude and the length of the path.

A particle oscillates simple harmonic motion with a period of 16 s . Two second after crossing the equilibrium position its velocity becomes 1 m//s . The amplitude is

A particle oscillates simple harmonic motion with a period of 16 s . Two second after crossing the equilibrium position its velocity becomes 1 m//s . The amplitude is

A particle executes SHM, its time period is 16 s. If it passes through the centre of oscillation, then its velocity is 2 ms^(-1) at times 2 s. The amplitude will be

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 3.14 cm//s . The frequency of its oscillation is

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm//s . The frequency of its oscillation is

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cm / s . The frequency of its oscillation is