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
- A \(N^2+1\)
- B \(\frac{1}{N^2}\)
- C \(N^2\)
- D \(N^2-1\)
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…
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