MHT CET · Physics · Oscillations
Two particles P and Q performs S.H.M. of same amplitude and frequency along the same straight line. At a particular instant, maximum distance between two particles is \(\sqrt{2} \mathrm{a}\). The initial phase difference between them is
- A \(\frac{\pi}{6}\)
- B \(\frac{\pi}{2}\)
- C zero
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
Maximum distance is \(\sqrt{2} \mathrm{a}\), means one particle lies at \(\mathrm{x}=\frac{\mathrm{a}}{\sqrt{2}}\) and second lies at \(\mathrm{x}=-\frac{\mathrm{a}}{\sqrt{2}}\).
Hence, \(x_1=a \sin \left(\omega t+\frac{\pi}{4}\right)\) and \(x_2=a \sin \left(\omega t-\frac{\pi}{4}\right)\)
Thus initial phase difference between the particles is \(\frac{\pi}{2}\)
Hence, \(x_1=a \sin \left(\omega t+\frac{\pi}{4}\right)\) and \(x_2=a \sin \left(\omega t-\frac{\pi}{4}\right)\)
Thus initial phase difference between the particles is \(\frac{\pi}{2}\)
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