The time period of a simple pendulum of lengt | vedclass

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(B) For a simple pendulum of length $L$,the time period is given by $T_1 = 2 \pi \sqrt{\frac{L}{g}}$.For a uniform rod of length $L$ suspended from on

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$\sqrt{\frac{3}{2}}$

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(B) For a simple pendulum of length $L$,the time period is given by $T_1 = 2 \pi \sqrt{\frac{L}{g}}$.For a uniform rod of length $L$ suspended from one end,the moment of inertia about the pivot is $I = \frac{1}{3} mL^2$. The distance of the center of mass from the pivot is $r_{cm} = \frac{L}{2}$.The time period of a physical pendulum is $T = 2 \pi \sqrt{\frac{I}{mg r_{cm}}}$.Substituting the values for the rod: $T_2 = 2 \pi \sqrt{\frac{\frac{1}{3} mL^2}{mg (L/2)}} = 2 \pi \sqrt{\frac{2L}{3g}}$.Now,calculating the ratio $\frac{T_1}{T_2} = \frac{2 \pi \sqrt{L/g}}{2 \pi \sqrt{2L/3g}} = \sqrt{\frac{L/g}{2L/3g}} = \sqrt{\frac{3}{2}}$.

The time period of a simple pendulum of length $L$ is $T_1$. The time period of a uniform rod of the same length $L$ suspended from one end and oscillating in a vertical plane is $T_2$. The amplitude of oscillation is small in both cases. Then the ratio $\frac{T_1}{T_2}$ is:

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