Find the amplitude and initial phase of a partical in SHM, whose motion equation is given as
`y = A sin omega t + B cos omega t`

The displacement of a particle executing simple harmonic motion is given by y= A_(0) +A sin omega t+ B cos omega t . Then the amplitude of its oscillation is given by:

The displacement equation of a particle of medium during wave mation is given by y=A sin omega t-B cos omega t The amplitude of the oscillator will be

The motion of a particle is given x = A sin omega t + B cos omega t The motion of the particle is

Whatis the displacement equation of S.H.M. with an amplitude 2m , if 120 oscillations are performed during one minute and initial phase is 60^(@) [ Consider displacement time equation of the form y = A sin( omega t + phi)] ?

Whatis the displacement equation of S.H.M. with an amplitude 2m , if 120 oscillations are performed during one minute and initial phase is 60^(@) [ Consider displacement time equation of the form y = A sin( omega t + phi)] ?

Find the resultant amplitude of the following simple harmonic equations : x_(1) = 5sin omega t x_(2) = 5 sin (omega t + 53^(@)) x_(3) = - 10 cos omega t

The SHM of a particle is given by the equation x=2 sin omega t + 4 cos omega t . Its amplitude of oscillation is

The SHM of a particle is given by the equation x=2 sin omega t + 4 cos omega t . Its amplitude of oscillation is

The motion of a particle is given by x = A sin omega t + B os omega t . The motion of the particle is