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

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 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 (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

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

Graph between velocity and displacement of a particle, executing S.H.M. is

If displacement x and velocity v related as 4v^(2) = 25 - x^(2)m in a SHM Then time period of given SHM is (consider SI unit)

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

If the displacement of a particle executing S.H.M. is given by x = 0.24 sin (400 t + 0.5)m, then the maximum velocity of the particle is