Y= Asin(`omega`t + `phi`) is the time - displacement equation of SHM , at t=0 the displacement of the particle is `y=A/2` and it is moving along negative x - direction .Then the initial phase angle `phi` will be

`2pi``/3`

`pi`/3

`pi`/6

5`pi`/6

The displacement-time equation of a particle executing SHM is A-Asin(omegat+phi) . At tme t=0 position of the particle is x= A/2 and it is moving along negative x-direction. Then the angle phi can be

The displacement of a particle is given by x = 3 sin ( 5 pi t) + 4 cos ( 5 pi t) . The amplitude of particle is

The displacement of a particle is represented by the equation y=3cos((pi)/(4)-2omegat). The motion of the particle is

The displacement of a particle is represented by the equation y=3cos((pi)/(4)-2omegat). The motion of the particle is

The displacement of a particle is represented by the equation y=3cos((pi)/(4)-2omegat). The motion of the particle is

The displacement of a particle is repersented by the equation y=3cos((pi)/(4)-2omegat) . The motion of the particle is (b) (c) (d)

Displacement of a particle executing SHM s x= 10 ( cos pi t + sin pi t) . Its maximum speed is

A particle is executing SHM given by x = A sin (pit + phi) . The initial displacement of particle is 1 cm and its initial velocity is pi cm//sec . Find the amplitude of motion and initial phase of the particle.

The phase of a particle in SHM at time t is (13pi)/(6) . The following interference is drawn form this

The motion of a particle in S.H.M. is described by the displacement function, x=Acos(omegat+phi) , If the initial (t=0) position of the particle is 1cm and its initial velocity is omega cm s^(-1) , what are its amplitude and initial phase angle ? The angular frequency of the particle is pis^(-1) . If instead of the cosine function, we choose the sine function to describe the SHM : x=B sin(omegat+alpha) , what are the amplitude and initial phase of the particle with the above initial conditions ?