A partical excuting SHM oscillates be between two fixed points separeted by `20 cm` . If its maximum velocity be `30 cm//s`. Find its velocity when its displacement is `5 cm` from its mean position.

A particle executing SHM oscillates between two fixed points separated by 20 cm . If its maximum velocity be 30 cm//s . Find its velocity when its displacement is 5 cm from its mean position.

A particle executing a SHM has maximum acceleration at a distance of 0.5 cm from its mean position is 2cm//s^(2) . What will be its velocity when it is at a distance of 1 cm from its mean position.

A particle is executing SHM with amplitude 4 cm and has a maximum velocity of 10 cm/sec What is its velocity at displacement 2 cm?

A particle is executing SHM with amplitude 4 cm and has a maximum velocity of 10 cm/sec What is its velocity at displacement 2 cm?

For Question , find maximum velocity and maximum acceleration Hint : Maximum velocity = Aomega Maximum acceleration = A omega^(2) A particle executes SHM on a straight line path. The amplitude of oscillation is 3cm . magnitude of its acceleration is equal to that of its velocity when its displacement from the mean position is 1cm.

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 body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. If its maximum velocity is 100 cm/s then its velocity will be 50 cm/s at a distance of

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)