MHT CET · Physics · Oscillations
A body performs linear simple harmonic motion of amplitude 'A". At what
displacement from the mean position, the potential energy of the body is one
fourth of its total energy?
- A \(\frac{\mathrm{A}}{3}\)
- B \(\frac{\mathrm{A}}{2}\)
- C \(\frac{3 \mathrm{~A}}{4}\)
- D \(\frac{A}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{A}}{2}\)
Step-by-step Solution
Detailed explanation
Total energy \(=\frac{1}{2} \mathrm{KA}^{2}\)
Potential energy \(=\frac{1}{2} \mathrm{kx}^{2}\)
If P.E. \(=\frac{1}{4}(\mathrm{~K} . \mathrm{E} .)\) then
\(\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{4}\left(\frac{1}{2} \mathrm{kA}^{2}\right)\)
\(\therefore x^{2}=\frac{A^{2}}{4}\) or \(x=\frac{A}{2}\)
Potential energy \(=\frac{1}{2} \mathrm{kx}^{2}\)
If P.E. \(=\frac{1}{4}(\mathrm{~K} . \mathrm{E} .)\) then
\(\frac{1}{2} \mathrm{kx}^{2}=\frac{1}{4}\left(\frac{1}{2} \mathrm{kA}^{2}\right)\)
\(\therefore x^{2}=\frac{A^{2}}{4}\) or \(x=\frac{A}{2}\)
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