A pendulum of length L oscillates with an angular amplitude of = 60^ and time period T. Let T_0 = 2 L g be the time period for small angle of oscillations, where g is the acceleration due to gravity. If air resistance is negligibly small and the string remains straight, then which one of the following is correct?
A. T will be slightly greater than T_0.
The restoring torque acting on the pendulum is = -mgL . The angular acceleration is = - g L . For small angular amplitudes, , which gives the time period T_0 = 2 L g . For a large angular amplitude ( = 60^ ), Due to the smaller restoring torque, the angular acceleration is reduced, which increases the time taken to complete an oscillation. The exact time period for large amplitudes is given by the series expansion T = T_0 ( 1 + 1 4 ^2 ( _0 2 ) + ). Since _0 = 60^ , the terms in the bracket are positive, yielding T > T_0. Therefore, T will be slightly greater than T_0. Answer: T will be slightly greater than T_0.
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