MHT CET · Physics · Oscillations
A simple pendulum of length \(l_1\) has time period \(T_1\). Another simple pendulum of length \(l_2\left(l_1 \gt l_2\right)\) has time period \(T_2\). Then the time period of the pendulum of length \(\left(l_1-l_2\right)\) will be
- A \(T_1-T_2\)
- B \(\sqrt{\frac{T_1}{T_2}}\)
- C \(\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2}\)
- D \(\sqrt{\frac{T_2}{T_1}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2}\)
Step-by-step Solution
Detailed explanation
For a simple pendulum,
\(\mathrm{T}_1=2 \pi \sqrt{\frac{l_1}{\mathrm{~g}}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{l_2}{\mathrm{~g}}}\)
Now, \(\mathrm{T}_1{ }^2=4 \pi \times \frac{l_1}{\mathrm{~g}}\) and \(\mathrm{T}_2^2=4 \pi \times \frac{l_2}{\mathrm{~g}}\)
\(\therefore \quad l_1=\frac{\mathrm{T}_1^2 \mathrm{~g}}{4 \pi}\) and \(l_2=\frac{\mathrm{T}_2^2 \mathrm{~g}}{4 \pi}\)
\(\therefore \quad l_1-l_2=\left(\mathrm{T}_1^2-\mathrm{T}_2^2\right) \times \frac{\mathrm{g}}{4 \pi}\)
Time period of pendulum of length \(\left(l_1-l_2\right)\) is,
\(\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\left(l_1-l_2\right)}{\mathrm{g}}}=2 \pi \sqrt{\frac{\left(\mathrm{~T}_1^2-\mathrm{T}_2^2\right) \times \frac{\mathrm{g}}{4 \pi}}{\mathrm{~g}}} \\
& \therefore \quad \mathrm{~T}=\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2}
\end{aligned}\)
\(\mathrm{T}_1=2 \pi \sqrt{\frac{l_1}{\mathrm{~g}}} \text { and } \mathrm{T}_2=2 \pi \sqrt{\frac{l_2}{\mathrm{~g}}}\)
Now, \(\mathrm{T}_1{ }^2=4 \pi \times \frac{l_1}{\mathrm{~g}}\) and \(\mathrm{T}_2^2=4 \pi \times \frac{l_2}{\mathrm{~g}}\)
\(\therefore \quad l_1=\frac{\mathrm{T}_1^2 \mathrm{~g}}{4 \pi}\) and \(l_2=\frac{\mathrm{T}_2^2 \mathrm{~g}}{4 \pi}\)
\(\therefore \quad l_1-l_2=\left(\mathrm{T}_1^2-\mathrm{T}_2^2\right) \times \frac{\mathrm{g}}{4 \pi}\)
Time period of pendulum of length \(\left(l_1-l_2\right)\) is,
\(\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{\left(l_1-l_2\right)}{\mathrm{g}}}=2 \pi \sqrt{\frac{\left(\mathrm{~T}_1^2-\mathrm{T}_2^2\right) \times \frac{\mathrm{g}}{4 \pi}}{\mathrm{~g}}} \\
& \therefore \quad \mathrm{~T}=\sqrt{\mathrm{T}_1^2-\mathrm{T}_2^2}
\end{aligned}\)
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