MHT CET · Physics · Oscillations
The period of S.H.M. of a particle is 16 second. The phase difference between the positions at \(\mathrm{t}=2 \mathrm{~s}\) and \(\mathrm{t}=4 \mathrm{~s}\) will be
- A \(\pi\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{4}\)
- D \(\frac{\pi}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{4}\)
Step-by-step Solution
Detailed explanation
\(\omega = \frac{2\pi}{T} = \frac{2\pi}{16} = \frac{\pi}{8}\) \(\Delta t = t_2 - t_1 = 4 - 2 = 2 \text{ s}\)
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