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Question 14

Choose the correct length $$(L)$$ versus square of time period $$(T_2)$$ graph for a simple pendulum executing simple harmonic motion

Solution

From simple pendulum relation

$$T=2\pi\sqrt{\frac{L}{g}}$$

Squaring,

$$T^2=\frac{4\pi^2L}{g}$$

Rearrange:

$$\frac{1}{T^2}=\frac{g}{4\pi^2L}$$

So

$$T^2\propto L$$

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