MHT CET · Physics · Oscillations
A pendulum is oscillating with frequency 'n' on the surface of earth. If it is taken to
depth \(\frac{\mathrm{R}}{2}\) below the surface of earth where \(\mathrm{R}\) is radius of earth. New frequency of
oscillations at depth \(\frac{\mathrm{R}}{2}\) is
- A \(\frac{n}{\sqrt{2}}\)
- B \(\mathbf{n}\)
- C \(\frac{\mathrm{n}}{\sqrt{3}}\)
- D \(2 n\)
Answer & Solution
Correct Answer
(A) \(\frac{n}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} T=2 \pi \sqrt{\frac{l}{g}} \quad T^{\prime} &=2 \pi \sqrt{\frac{1}{g^{\prime}}} \\ &=2 \pi \sqrt{\frac{1}{g / 2}} \\ T^{\prime} &=\frac{\sqrt{2}}{T} \\ \therefore n^{\prime} &=\frac{n}{\sqrt{2}} \end{aligned}\)
\(n=1 / T\)
\(\begin{aligned} g^{\prime} &=g(1-d / R) \\ &=g\left(1-\frac{R / 2}{R}\right) \\ &=g / 2 \end{aligned}\)
\(n=1 / T\)
\(\begin{aligned} g^{\prime} &=g(1-d / R) \\ &=g\left(1-\frac{R / 2}{R}\right) \\ &=g / 2 \end{aligned}\)
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