MHT CET · Physics · Oscillations

A particle performs S.H.M. from the mean position. Its amplitude is 'A' and total
energy is 'E'. At a perticular instant its kinetic energy is \(\frac{3 \mathrm{E}}{4}\). The displacement of the
particle at that instant is

A \(A\)
B \(\frac{A}{8}\)
C \(\frac{A}{4}\)
D \(\frac{A}{2}\)

Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{A}{2}\)

Step-by-step Solution

Detailed explanation

Total energy \(=\mathrm{E}=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\)
When kinetic energy is \(\frac{3 \mathrm{E}}{4}\), its potential energy is \(\left(\mathrm{E}-\frac{3 \mathrm{E}}{4}\right)=\frac{\mathrm{E}}{4}\)
Dividing Eq.(1) by Eq.(2), \(\quad 4=\frac{A^{2}}{x^{2}} \quad \therefore x=\frac{A}{2}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A particle performs S.H.M. from the mean position. Its amplitude is 'A' and totalenergy is 'E'. At a perticular instant its kinetic energy is \(\frac{3 \mathrm{E}}{4}\). The displacement of theparticle at that instant is

Step-by-step Solution

A particle performs S.H.M. from the mean position. Its amplitude is 'A' and total
energy is 'E'. At a perticular instant its kinetic energy is \(\frac{3 \mathrm{E}}{4}\). The displacement of the
particle at that instant is