TS EAMCET · Physics · Oscillations

Two particles execute simple harmonic motion (SHM) along close parallel lines. SHM of the both the particles have same frequency and same amplitude. When they pass each other moving in opposite direction each time, their displacement is half their amplitude. Then, their phase difference is

A 0
B \(2 \pi / 3\)
C \(\pi / 3\)
D \(\pi / 2\)

Verified Solution

Answer & Solution

Correct Answer

(B) \(2 \pi / 3\)

Step-by-step Solution

Detailed explanation

The two SHM have same amplitude and same frequency. \[ \begin{aligned} \therefore \text { Let } x_1 & =\mathrm{a} \sin \omega t \\ & \text { and } x_2=\mathrm{a} \sin (\omega t+\phi) \end{aligned} \] (where, \(\phi\) is the phase difference.) When they cross each other,…

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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