The equation of motion of a particle executing SHM is `((d^2 x)/(dt^2))+kx=0`. The time period of the particle will be :

The equation of motion of a particle executing SHM is 3(d^(2)x)/(dt^(2))=27x=0 , what will be the angular frequency of the SHM?

The equation of motion of a particle executing S.H.M. is a = - bx , where a is the acceleration of the particle, x is the displacement from the mean position and b is a constant. What is the time period of the particle ?

The equation of motion of a particle executing SHM is (k is a positive constant)

The differential equation of a particle performing a S.H.M. is (d^(2)x)/(dt^(2))+ 64x=0 . The period of oscillation of the particle is

The equation of motion of a particle in SHM is a +16 pi^(2)x = 0 . Here 'a' is linear acceleration of the particle at displacement x (a,x are in SI) . Its time period is

The equation of motion of a particle in SHM is a +4x = 0 . Here 'a' is linear acceleration of the particle at displacement 'x' in metre. Its time period is

The equation of simple harmonic motion of a source is (d^2x)/(dt^2)+px=0 , find the time period.

The equation of a particle executing SHM is (d^(2)x)/(dt^(2))=-omega^(2)x . Where omega=(2pi)/("time period") . The velocity of particle is maximum when it passes through mean position and its accleration is maximum at extremeposition. The displacement of particle is given by x=A sin(omegat+theta) where theta -initial phase of motion. A -Amplitude of motion and T-Time period The time period of pendulum is given by the equation T=2pisqrt((l)/(g)) . Here (d^(2)x)/(dt^(2)) is :