A particle moves such that acceleration is given by `a= -4x`. The period of oscillation is

A particle moves as such acceleration is given by a = 3 sin 4t, then :

A particle moves such that its acceleration a is given by a = -bx , where x is the displacement from equilibrium positionand is a constant. The period of oscillation is

A particle moves such that its acceleration ‘a’ is given by a = – zx where x is the displacement from equilibrium position and z is constant. The period of oscillation is

A particle moves such that its acceleration is given by a = -Beta(x-2) Here, Beta is a positive constnt and x the x-coordinate with respect to the origin. Time period of oscillation is

A particle moves such that its acceleration is given by a =- beta (x-2) Here beta is positive constant and x is the position form origin. Time period of oscillation is

A particle moves along X axis such that its acceleration is given by a = -beta(x -2) ,where beta is a positive constant and x is the position co-ordinate. (a) Is the motion simple harmonic? (b) Calculate the time period of oscillations. (c) How far is the origin of co-ordinate system from the equilibrium position?

A particle is moving along the x-axis whose acceleration is given by a= 3x-4 , where x is the location of the particle. At t = 0, the particle is at rest at x = 4//3m . The distance travelled by the particles in 5 s is

A particle moves in such a way that its acceleration a= - bx, where x is its displacement from the mean position and b is a constant. The period of its oscillation is