MHT CET · Physics · Oscillations
The ratio of frequencies of two oscillating pendulums are \(3: 2\). Their lengths are in the ratio
- A \(\sqrt{2}: \sqrt{3}\)
- B \(9: 4\)
- C \(\sqrt{3}: \sqrt{2}\)
- D \(4: 9\)
Answer & Solution
Correct Answer
(D) \(4: 9\)
Step-by-step Solution
Detailed explanation
The frequency of oscillation is given by:
\(\mathrm{n}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\)
So, Frequency \(\mathrm{n} \propto \frac{1}{\sqrt{\mathrm{L}}}\)
\(\Rightarrow \frac{\mathrm{n}_1}{\mathrm{n}_2}=\sqrt{\frac{\mathrm{L}_2}{\mathrm{~L}_1}}\)
\(\Rightarrow \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{n}_2^2}{\mathrm{n}_1^2}=\frac{4}{9}\)
\(\mathrm{n}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\mathrm{L}}}\)
So, Frequency \(\mathrm{n} \propto \frac{1}{\sqrt{\mathrm{L}}}\)
\(\Rightarrow \frac{\mathrm{n}_1}{\mathrm{n}_2}=\sqrt{\frac{\mathrm{L}_2}{\mathrm{~L}_1}}\)
\(\Rightarrow \frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{n}_2^2}{\mathrm{n}_1^2}=\frac{4}{9}\)
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