A pendulum of length L oscillates with an angular amplitude of \theta = 60^\circ and time period T. Let T_0 = 2\pi \sqrt{\frac{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. ✓
- B. T will be slightly smaller than T_0.
- C. T will be exactly equal to T_0.
- D. T will depend upon the mass of the bob.
Correct Answer: A. T will be slightly greater than T_0.
Explanation
The time period of a simple pendulum increases with the angular amplitude. For large amplitudes such as 60 degrees, the approximation \sin\theta \approx \theta does not hold, leading to a longer time period T > T_0.
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