AP EAMCET · PHYSICS · Oscillations
A particle is executing simple harmonic motion with an instantaneous displacement \(x=A \sin ^2\left(\omega t-\frac{\pi}{4}\right)\). The time period of oscillation of the particle is
- A \(\frac{2 \pi}{\omega}\)
- B \(\frac{\pi}{\omega}\)
- C \(\frac{\pi}{2 \omega}\)
- D \(\frac{\omega}{2 \pi}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{\omega}\)
Step-by-step Solution
Detailed explanation
Given, instantaneous displacement of particle executing SHM. \(x=A \sin ^2(\omega t-4)\) \(\Rightarrow \quad x=A\left[\frac{1-\cos 2(\omega t-4)}{2}\right]\) \(\left(\right.\) Since, \(\left.\sin ^2 \theta=\frac{1-\cos 2 \theta}{2}\right)\)…
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