The potential energy of a particle oscillating on x-axis is given as `U +20 +(x-2)^(2)`. The mean position is at

The potential energy of a particle oscillating on x-axis is given as U = 20 + (x-2)^(2) where .U. is in joules and .x. is in metres. Its mean position is

The potential energy of a particle oscillating along x-axis is given as U=20+(x-2)^(2) Here, U is in joules and x in meters. Total mechanical energy of the particle is 36J . (a) State whether the motion of the particle is simple harmonic or not. (b) Find the mean position. (c) Find the maximum kinetic energy of the particle.

The potential energy of a particle oscillating along x-axis is given as U=20+(x-2)^(2) Here, U is in joules and x in meters. Total mechanical energy of the particle is 36J . (a) State whether the motion of the particle is simple harmonic or not. (b) Find the mean position. (c) Find the maximum kinetic energy of the particle.

The potential energy of a particle oscillating along x-axis is given as U=20+(x-2)^(2) Here, U is in joules and x in meters. Total mechanical energy of the particle is 36J . (a) State whether the motion of the particle is simple harmonic or not. (b) Find the mean position. (c) Find the maximum kinetic energy of the particle.

The potential energy U of a particle oscillating along x-axis is given as U=10-20x+5x^2 . If it is released at x=-2 , then maximum value of x is

The potential energy of a particle moving along x-axis is given by U = 20 + 5 sin (4 pi x) , where U is in J and x is in metre under the action of conservative force :

The potential energy of a particle in motion along X axis is given by U = U_0 - u_0 cos ax. The time period of small oscillation is

The potential energy U(x) of a particle moving along x - axis is given by U(x)=ax-bx^(2) . Find the equilibrium position of particle.

The potential energy U(x) of a particle moving along x - axis is given by U(x)=ax-bx^(2) . Find the equilibrium position of particle.