Physics - SHM Question with Solution | TestHub

The time period of a simple pendulum of length $l$ is given by $T=2\pi \sqrt{\frac{l}{g}}$.

Squaring both sides, we get $T^2=\frac{4{\pi }^{2}l}{g}$, which gives $l=\frac{g{T}^2}{4{\pi }^2}$.

For the first pendulum, $l_1=\frac{g{T}_1^2}{4{\pi }^2}$.

For the second pendulum, $l_2=\frac{g{T}_2^2}{4{\pi }^2}$.

The length of the new pendulum is $l=l_1-l_2$.

Substituting the values of $l_1$ and $l_2$, we get $l=\frac{g}{4{\pi }^2}(T_1^2-T_2^2)$.

The time period of the new pendulum is $T=2\pi \sqrt{\frac{l}{g}}$.

Substituting the value of $l$, we get $T=2\pi \sqrt{\frac{\frac{g}{4{\pi }^2}(T_1^2-T_2^2)}{g}}=\sqrt{T_1^2-T_2^2}$.

Answer: $\sqrt{{\mathrm{T}}_1^2-{\mathrm{T}}_2^2}$