Question 286: Oscillations

Question

A simple pendulum of length $l_1$ has time period $\mathrm{T}_1$. Another simple pendulum of length $l_2\left(l_1>l_2\right)$ has time period $T_2$. Then the time period of the pendulum of length $\left(l_1-l_2\right)$ will be

Accepted Answer

For a simple pendulum,
$$\mathrm{T}_1=2 \pi \sqrt{\frac{l_1}{\mathrm{~g}}} 	ext { and } \mathrm{T}_2=2 \pi \sqrt{\frac{l_2}{\mathrm{~g}}}$$
Now, $\mathrm{T}_1^2=4 \pi 	imes \frac{l_1}{\mathrm{~g}}$ and $\mathrm{T}_2^2=4 \pi 	imes \frac{l_2}{\mathrm{~g}}$
$$\begin{array}{ll}
	herefore & l_1=\frac{\mathrm{T}_1^2 \mathrm{~g}}{4 \pi} 	ext { and } l_2=\frac{\mathrm{T}_2^2 \mathrm{~g}}{4 \pi} \
	herefore & l_1-l_2=\left(\mathrm{T}_1^2-\mathrm{T}_2^2ight) 	imes \frac{\mathrm{g}}{4 \pi}\quad	ext{.... (i)}
\end{array}$$
Time period of pendulum of length $\left(l_1-l_2ight)$ is,
$$\begin{aligned}
& \mathrm{T} =  2 \pi \sqrt{\frac{\left(l_1-l_2ight)}{\mathrm{g}}}=2 \pi \sqrt{\frac{\left(\mathrm{~T}_1^2-\mathrm{T}_2^2ight) 	imes \frac{\mathrm{g}}{4 \pi}}{\mathrm{~g}}} \
	herefore \quad \mathrm{~T} & =\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2}
\end{aligned}$$

Answer

$T_1-T_2$

Answer

$\sqrt{\frac{T_1}{T_2}}$

Answer

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

Answer

$\sqrt{\frac{T_2}{T_1}}$

Oscillations

A simple pendulum of length l_1 has time period \mathrm{T}_1. Another simple pendulum of length l_2\left(l_1>l_2\right) has time period T_2. Then the time period of the pendulum of length \left(l_1-l_2\right) will be

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