As illustrated in the figure, a small block undergoes back-and-forth oscillations on a smooth concave surface of radius R . Determine the time period for small oscillations.

Step-by-step solution

Let the block be displaced by a small angle from the lowest point (mean position). The restoring force acting on the block along the tangential direction is: F = -mg For small oscillations, . Thus, the restoring force becomes: F = -mg The displacement along the arc is x = R , which implies = x R . Substituting this into the force equation gives: F = -mg ( x R ) The acceleration of the block is: a = F m = - ( g R ) x Comparing this with the standard equation of simple harmonic motion a = - ^2 x , we get: ^2 = g R =