MHT CET · Physics · Oscillations
The period of a simple pendulum gets doubled when
- A its length is doubled.
- B its length is made four times.
- C its length is made half.
- D the mass of the bob is doubled.
Answer & Solution
Correct Answer
(B) its length is made four times.
Step-by-step Solution
Detailed explanation
We know for a simple pendulum,
\(\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \Rightarrow \mathrm{~T} \propto \sqrt{l} \\
& \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\sqrt{\frac{l_1}{l_2}}
\end{aligned}\)
When \(l_2=4 l_1\),
\(\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\sqrt{\frac{l_1}{4 l_1}} \Rightarrow \mathrm{~T}_2=2 \mathrm{~T}_1\)
\(\therefore \quad\) Period is doubled when length is made four times
\(\begin{aligned}
& \frac{\mathrm{T}_1}{2 \mathrm{~T}_1}=\sqrt{\frac{l_1}{l_2}} \\
& \frac{1}{2}=\sqrt{\frac{l_1}{l_2}} \Rightarrow \frac{1}{4}=\frac{l_1}{l_2} \Rightarrow l_2=4 l_1
\end{aligned}\)
\(\therefore \quad\) Period is doubled when length is made four times
\(\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \Rightarrow \mathrm{~T} \propto \sqrt{l} \\
& \frac{\mathrm{~T}_1}{\mathrm{~T}_2}=\sqrt{\frac{l_1}{l_2}}
\end{aligned}\)
When \(l_2=4 l_1\),
\(\frac{\mathrm{T}_1}{\mathrm{~T}_2}=\sqrt{\frac{l_1}{4 l_1}} \Rightarrow \mathrm{~T}_2=2 \mathrm{~T}_1\)
\(\therefore \quad\) Period is doubled when length is made four times
\(\begin{aligned}
& \frac{\mathrm{T}_1}{2 \mathrm{~T}_1}=\sqrt{\frac{l_1}{l_2}} \\
& \frac{1}{2}=\sqrt{\frac{l_1}{l_2}} \Rightarrow \frac{1}{4}=\frac{l_1}{l_2} \Rightarrow l_2=4 l_1
\end{aligned}\)
\(\therefore \quad\) Period is doubled when length is made four times
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