Equation of S.H.M.is :

The equation of S.H.M. with amplitude 4m and time period 1/2 s with initial phase pi//3 is x =

The equation of S.H.M. of a particle is a+4pi^(2)x=0 , where a is instantaneous linear acceleration at displacement x. Then the frequency of motion is

The amplitude and the time period in a S.H.M. is 0.5 cm and 0.4 sec respectively. If the initial phase is pi//2 radian, then the equation of S.H.M. will be

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)] ?

Obtain the equation of S.H.M. of a particle whose amplitude is 0.02,and whose frequency is 25 Hz. The initial phase is (pi)/(4) Hint : x= A sin ( omegat +phit) omega = 2pi v

If displacement time equation of an S.H.M. is x= sin . (pit)/(6) , then the moment of times at which peak value of acceleration is attained are

The equation of a S.H.M. of amplitude 'A' and angular frequency omega in which all distances are measured from one extreme positionand time is taken to be zero, at the other extreme position is

Equation of two S.H.M. x_(1)=5 sin (2pi t+pi//4),x_(2)=5sqrt(2)(sin 2pit+cos 2pit) ratio of amplitude & phase difference will be

The equation of S.H.M of a particle whose amplitude is 2 m and frequency 50 Hz. Start from extreme position is

Write the equation of an S.H.M of amplitude 5 cm if 150 oscillations are performed in one minute and the intial phase is 45^(@)