Simple Harmonic...

An S.H.M. has an amplitude ‘a’ and  a time period T. The maximum velocity will be -

Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is -

The angular velocities of three bodies in simple harmonic motion are ω₁,ω_2,ω₃ with their respective amplitudes as A_1,A₂,A₃. If all the three bodies have same mass and maximum velocity, then

The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes 8√3 cm/sec will be

The maximum velocity of simple harmonic motion represented by y = 3sin (100t+6/π​) is given by

The displacement equation of a particle is x=3 sin 2t + 4cos 2t The amplitude and maximum velocity will be respectively

The instantaneous displacement of a simple pendulum oscillator is given by x = A cos (ωt + π/4) . Its speed will be maximum at time

The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by

What is the maximum acceleration of the particle doing the SHM  where 2 is in cm