The potential energy of a particle of mass `1kg` in motion along the x- axis is given by: `U = 4(1 - cos 2x)`, where `x` in meters. The period of small oscillation (in sec) is

The potential energy of a particle of mass 1kg in motion along the x- axis is given by: U = 4(1 - cos 2x) , where x in metres. The period of small oscillation (in sec) is

The potential energy of a particle of mass 1 kg in motin along the x-axis is given by U = 4(1-cos2x)J Here, x is in meter. The period of small osciallationis (in second) is

The potential energt of a particle of mass 0.1 kg, moving along the x-axis, is given by U=5x(x-4)J , where x is in meter. It can be concluded that

The potential energy of particle of mass 1kg moving along the x-axis is given by U(x) = 16(x^(2) - 2x) J, where x is in meter. Its speed at x=1 m is 2 m//s . Then,

The potential energy of body of mass 2 kg moving along the x-axis is given by U = 4x^2 , where x is in metre. Then the time period of body (in second) is

The potential energy of a particle of mass 2 kg moving along the x-axis is given by U(x) = 4x^2 - 2x^3 ( where U is in joules and x is in meters). The kinetic energy of the particle is maximum at

The potential energy of a particle of mass m is given by U(x)=U_0(1-cos cx) where U_0 and c are constants. Find the time period of small oscillations of the particle.

The potential energy of a particle of mass 1 kg moving along x-axis given by U(x)=[(x^(2))/(2)-x]J . If total mechanical speed (in m/s):-

The potential energy of a particle of mass 1 kg moving in X-Y plane is given by U=(12x+5y) joules, where x an y are in meters. If the particle is initially at rest at origin, then select incorrect alternative :-