The potential energy of a particle in motion along x axis is experienced by `uu=uu_0-uu_0cos ax` Find the time period of the particle.

The potential energy of a particle of mass 0.02 kg moving along x -axis is given by V = Ax(x - 4) J where x is in metres and A is a constant. Which of the following is/are correct statement (s)? (A) The particle is acted upon by a constant force. (B) The particle executes simple harmonic motion. (C) The speed of the particle is maximum at x = 2 m. (D) The period of oscillation of the particle is pi/5 sec.

Potential energy of a particle is 1/2momega^(2)x^(2) , where m and omega are constants. Prove that the motion of the particle is simple harmonic.

The potential energy of a particle is given by the formula U= 100 -5x+100x^2 , U and x are in SI units. If mas of the particle is 0.1 kg then magnitude of its acceleration .

If the displacement and the acceleration of a particle executing SHM at any instant are x and a respectively, then the time period of that motion will be

Total energy of a particle executing simple harmonic motion is 400 erg and the maximum force acting on the particle is 100 dyn. If the time period is 2 s and the initial phase is 30^(@) , then write down the equation of motion. What is the mass of the particle?

A particle of mass m is located in a one dimensional potential field where potential energy is given by V(x) = A(1- cos Px) , where A and P are constant . The period of small oscillations of the particle is