What is the time period of a simple pendulum of length \( L \) and acceleration due to gravity \( g \)?
Show Hint
- \( 2\pi \sqrt{\frac{L}{g}} \)
- \( 2\pi \sqrt{\frac{g}{L}} \)
- \( \frac{2\pi}{\sqrt{L}} \)
- \( \frac{2\pi}{g} \)
The Correct Option is A
Solution and Explanation
The time period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}}, \]
where \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity.
Therefore, the correct answer is option (1).
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