MHT CET · Physics · Oscillations
The ratio of frequencies of oscillations of two simple pendulums is \(3: 4\), then their lengths are in the ratio
- A \(16: 9\)
- B \(9:16\)
- C \(\sqrt{3}: \sqrt{4}\)
- D \(\sqrt{4}: \sqrt{3}\)
Answer & Solution
Correct Answer
(A) \(16: 9\)
Step-by-step Solution
Detailed explanation
Frequency of simple pendulum is given by
\(\begin{aligned}
\mathrm{f} &=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \quad \therefore \frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\sqrt{\frac{\ell_{2}}{\ell_{1}}} \\
\therefore \frac{3}{4} &=\sqrt{\frac{\ell_{2}}{\ell_{1}}} \quad \text { or } \quad \frac{9}{16}=\frac{\ell_{2}}{\ell_{1}} \\
\therefore \frac{\ell_{1}}{\ell_{2}} &=\frac{16}{9}
\end{aligned}\)
\(\begin{aligned}
\mathrm{f} &=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \quad \therefore \frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\sqrt{\frac{\ell_{2}}{\ell_{1}}} \\
\therefore \frac{3}{4} &=\sqrt{\frac{\ell_{2}}{\ell_{1}}} \quad \text { or } \quad \frac{9}{16}=\frac{\ell_{2}}{\ell_{1}} \\
\therefore \frac{\ell_{1}}{\ell_{2}} &=\frac{16}{9}
\end{aligned}\)
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