A simple pendulum performing small oscillations at a height R above Earth's surface has a time period of \(T_1 = 4\) s. What would be its time period at a point which is at a height \(2R\) from Earth's surface?

Question

The variation of displacement with time of a simple harmonic motion (SHM) for a particle of mass \( m \) is represented by: \[ y = 2 \sin \left( \frac{\pi}{2} + \phi \right) \, \text{cm} \] The maximum acceleration of the particle is:

Answer 1

The formula for the period of a simple pendulum is \( T = 2\pi \sqrt{\frac{L}{g}} \). Here, \(L\) represents the pendulum's length, and \(g\) denotes the acceleration due to gravity. While gravity decreases slightly at an altitude of \(R\) above Earth's surface, for small height variations, the period is expected to be approximately constant even at \(2R\). Therefore, \(T_1 = T_2\).

A simple pendulum performing small oscillations at a height R above Earth's surface has a time period of \(T_1 = 4\) s. What would be its time period at a point which is at a height \(2R\) from Earth's surface?

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The Correct Option is A

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