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AP EAMCET · PHYSICS · Oscillations

The time period of a simple pendulum on the surface of the earth is T. The height above the surface of the earth at which the time period of the pendulum becomes \(2 \mathrm{~T}\) is (Radius of the earth \(=6400 \mathrm{~km}\) )

  1. A \(3200 \mathrm{~km}\)
  2. B \(6400 \mathrm{~km}\)
  3. C \(1600 \mathrm{~km}\)
  4. D \(800 \mathrm{~km}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(3200 \mathrm{~km}\)

Step-by-step Solution

Detailed explanation

Time period of a simple pendulum is given by, \(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}\) \(2 \mathrm{~T}=\frac{\mathrm{T}}{1+\frac{\mathrm{h}}{\mathrm{R}}} \Rightarrow 1+\frac{\mathrm{h}}{\mathrm{R}}=\frac{1}{2}\)…
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