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TS EAMCET · Physics · Oscillations

The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of \(\frac{1}{N}\) of its amplitude from the mean position is

  1. A \(N^2+1\)
  2. B \(\frac{1}{N^2}\)
  3. C \(N^2\)
  4. D \(N^2-1\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(N^2-1\)

Step-by-step Solution

Detailed explanation

The kinetic energy \[ \mathrm{KE}=\frac{1}{2} m \omega^2\left[\mathrm{a}^2-\left(\frac{\mathrm{a}}{\mathrm{N}}\right)^2\right] \] The potential energy \[ \mathrm{PE}=\frac{1}{2} m \omega^2 \frac{\mathrm{a}^2}{\mathrm{~N}^2} \] From the Eqs. (i) and (ii), we get…