If the displacement (x) and velocity (v) of a particle executing SHM are related through the expression `3v^(2)=30-x^(2)`. If the time period of the particle is `T=pisqrt(n)`, then what is the value of n?

If the displacement (x) and velocity (v) of a particle executing simple harmonic motion are related through the expression 4v^2=25-x^2 , then its time period is given by

Dispalcement (x) and velocity (v) of a particle exectuting S.H.M are related as 2v^(2)=9-x^(2) . The time period of particle is

If the displacement x and velocity v of a aprticle executing SHM are related as 4v^(2) = 25 - x^(2) . Then its maximum displacement in metre (x, v are in SI ) is

If the displacement (y in m) and velocity (v in ms^(-1)) of a particle executing SHM are related by the equation 4v^(2)=16-y^(2) , then the path length of the motion and time period of oscillation respectively are

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 displacement (x) is plotted with time (t) for a particle executing SHM as shown below. The correct equation of SHM is

If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude the time period will be

If the maximum velocity and acceleration of a particle executing SHM are equal in magnitude, the time period will be

The plot of velocity (v) versus displacement (x) of a particle executing simple harmonic motion is shown in figure. The time period of oscillation of particle is :-