MHT CET · Physics · Oscillations
The instantaneous displacement of particle in S.H.M is \(x=A \cos \left(\omega t+\frac{\pi}{4}\right)\). The time at which the velocity is maximum for the first time is
- A \(\frac{\omega}{2 \pi}\)
- B \(\frac{\pi}{\omega}\)
- C \(\frac{2 \pi}{\omega}\)
- D \(\frac{\pi}{4 \omega}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi}{4 \omega}\)
Step-by-step Solution
Detailed explanation
Given, \(x=\operatorname{Acos}\left(\omega t+\frac{\pi}{4}\right)\)
\(\frac{\mathrm{dx}}{\mathrm{dt}}=(-\mathrm{A} \omega) \sin \left(\omega \mathrm{t}+\frac{\pi}{4}\right)\)
When \(\left(\omega t+\frac{\pi}{4}\right)=\frac{\pi}{2}, \sin \left(\omega t+\frac{\pi}{4}\right)=1\)
\(\therefore\) At time \(t=\frac{\pi}{4 \omega}\), the velocity is maximum for the first time
\(\frac{\mathrm{dx}}{\mathrm{dt}}=(-\mathrm{A} \omega) \sin \left(\omega \mathrm{t}+\frac{\pi}{4}\right)\)
When \(\left(\omega t+\frac{\pi}{4}\right)=\frac{\pi}{2}, \sin \left(\omega t+\frac{\pi}{4}\right)=1\)
\(\therefore\) At time \(t=\frac{\pi}{4 \omega}\), the velocity is maximum for the first time
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