AP EAMCET · PHYSICS · Oscillations
The equations for the displacements of two particles in simple harmonic motion are \(y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)\) and \(y_2=0.1 \cos \pi t\) respectively. The phase difference between the velocities of the two particles at a time \(t=0\) is
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{2}\)
- C \(\frac{\pi}{6}\)
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{6}\)
Step-by-step Solution
Detailed explanation
\(v_1 = \frac{dy_1}{dt} = 0.1(100\pi) \cos \left(100 \pi t+\frac{\pi}{3}\right) = 10\pi \cos \left(100 \pi t+\frac{\pi}{3}\right)\) \(v_2 = \frac{dy_2}{dt} = 0.1(-\pi) \sin \pi t = -0.1\pi \sin \pi t\) \(v_2 = 0.1\pi \cos \left(\pi t + \frac{\pi}{2}\right)\) Phase of \(v_1\) at…
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