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 such that its acceleration a is given by a = (-b)x, Then, the 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 moves such that acceleration is given by a= -4x . The period of oscillation is

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

A moving particle has an acceleration a = -bx, where b is a positive constant and x is the distance of the particle from the equilibrium position. What is the time period of oscillation of the particle?

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 accleration a is given by a = -bx , where x is the displacement from equilibrium position and b is a constant. The period of oscillation is