A particle executes simple harmonic motion with an angular velocity and maximum acceleration of `3.5rad//sec and 7.5 m//s^(2)` respectively. The amplitude of oscillation

A particle executes simple harmonic motion with an angular velocity and maximum acceleration of 3.5 rad // s and 7.5 m // s^(2) respectively. Amplitude of the oscillation is .......... .

A particle executes simple harmonic motion with an angular velocity and maximum acceleration of 3.5(rad)/(sec)" and "7.5(m)/(s^2) respectively. Amplitude of the oscillation is…………..

For a particle executing simple harmonic motion, the acceleration is -

If a particle executes SHM with angular velocity 3.5 rad/s and maximum acceleration 7.5m//s^(2) , then the amplitude of oscillation will be

If a particle is executing simple harmonic motion, then acceleration of particle

A particle executes a SHM of angular velocity 2 rad/s and maximum acceleration of 8 m//s^(2) . What is the path length of the oscillator ?

A particle is executing simple harmonic motion with an amplitude of 0.02 metre and frequency 50 Hz . The maximum acceleration of the particle is

For a particle executing simple harmonic motion, the acceleration is proportional to

For a particle executing simple harmonic motion, the acceleration is proportional to