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(B) The time period $T$ of a simple pendulum of length $L$ is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$.From this, we have $T^2 = 4\pi^2 \fra

Vedclass · Accepted Answer

$\sqrt{T_1^2 - T_2^2}$

Vedclass · Answer

(B) The time period $T$ of a simple pendulum of length $L$ is given by the formula $T = 2\pi \sqrt{\frac{L}{g}}$.From this, we have $T^2 = 4\pi^2 \frac{L}{g}$, which implies $L = \frac{T^2 g}{4\pi^2}$.For the two given pendulums:$L_1 = \frac{T_1^2 g}{4\pi^2}$ and $L_2 = \frac{T_2^2 g}{4\pi^2}$.We want to find the time period $T'$ for a pendulum of length $L' = L_1 - L_2$.Substituting the expressions for $L_1$ and $L_2$:$L' = \frac{T_1^2 g}{4\pi^2} - \frac{T_2^2 g}{4\pi^2} = \frac{g}{4\pi^2} (T_1^2 - T_2^2)$.Using the formula $T' = 2\pi \sqrt{\frac{L'}{g}}$:$T' = 2\pi \sqrt{\frac{g(T_1^2 - T_2^2)}{4\pi^2 g}} = 2\pi \sqrt{\frac{T_1^2 - T_2^2}{4\pi^2}} = \sqrt{T_1^2 - T_2^2}$.Thus, the correct option is $B$.

Two simple pendulums of lengths $L_1$ and $L_2$ have time periods $T_1$ and $T_2$ respectively, where $T_1 > T_2$. What is the time period of a simple pendulum of length $(L_1 - L_2)$?

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