Displacement-time equation of a particle executing SHM is `x=A sin (omega t+(pi)/6)` Time taken by the particle to go directly from `x=-A/2"to"x=+A/2`

Displacement-time equation of a particle execution SHM is x=A sin( omegat+pi/6) Time taken by the particle to go directly from x = -A/2 to x = + A/2 is

Displacement time equation of a particle executing SHM is, x = 10 sin ((pi)/(3)t+(pi)/(6))cm . The distance covered by particle in 3s is

The displecemen-time equation of a particle execitting SHM is x = A sin (omega t + phi) At time t = 0 position of are position is x = A//2 and it is moving along negative x- direction .Then the angle phi can be

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

Displacement-time graph of a particle executing SHM is as shown The corresponding force-time graph of the particle can be

Displacement-time equation of a particle executing SHM is x=4sin omega t+3tsin(omegat+pi//3) Here, x is in cm and t ini sec. The amplitude of oscillation of the particle is approximately.

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

The displacement equation of a particle performing S.H.M. is x = 10 sin (2pit+(pi)/(6)) m. Then the initial displacement of a particle is

A particle executes S.H.M with time period 12 s. The time taken by the particle to go directly from its mean position to half its amplitude.

x - t equation of a particle executing SHM is x = Acos (omega t - 45^(@)) Find the point from where particle starts its journey and the direction of its initial velocity.