Question 9

Choose the correct length ($$L$$) versus square of time period ($$T^2$$) graph for a simple pendulum executing simple harmonic motion.

Solution

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

$$T^2 = 4\pi^2 \left( \frac{L}{g} \right)$$

This equation is in the form $$y = mx$$, where $$y = T^2$$, $$x = L$$, and the slope $$m = \frac{4\pi^2}{g}$$. This is the equation of a straight line passing through the origin.

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Choose the correct length ($$L$$) versus square of time period ($$T^2$$) graph for a simple pendulum executing simple harmonic motion.

Solution

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