MHT CET · Physics · Gravitation
A pendulum is oscillating with frequency ' \(n\) ' on the surface of the earth. It is taken to a depth \(\frac{R}{2}\) below the surface of earth. New frequency of oscillation at depth \(\frac{\mathrm{R}}{2}\) is [ \(\mathrm{R}\) is the radius of earth]
- A \(\frac{\mathrm{n}}{3}\)
- B \(\frac{\mathrm{n}}{\sqrt{2}}\)
- C \(2 \mathrm{n}\)
- D \(\frac{\mathrm{n}}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{n}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Frequency of a simple pendulum is given by
\(
\begin{aligned}
& \mathrm{n}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \\
& \therefore \frac{\mathrm{n}^{\prime}}{\mathrm{n}}=\sqrt{\frac{\mathrm{g}^{\prime}}{\mathrm{g}}} \\
& \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2} \\
& \therefore \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\sqrt{2}}
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{n}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{g}}{\ell}} \\
& \therefore \frac{\mathrm{n}^{\prime}}{\mathrm{n}}=\sqrt{\frac{\mathrm{g}^{\prime}}{\mathrm{g}}} \\
& \mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R}}\right)=\mathrm{g}\left(1-\frac{1}{2}\right)=\frac{\mathrm{g}}{2} \\
& \therefore \mathrm{n}^{\prime}=\frac{\mathrm{n}}{\sqrt{2}}
\end{aligned}
\)
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