JEE Mains · Physics · STD 11 - 13. oscillations
The displacement of a particle executing SHM is given by \(x=10 \sin \left(\omega t+\frac{\pi}{3}\right) \mathrm{m}\). The time period of motion is \(3.14 \mathrm{~s}\). The velocity of the particle at \(\mathrm{t}=0\) is_________ \(\mathrm{m} / \mathrm{s}\).
- A \(10\)
- B \(15\)
- C \(20\)
- D \(25\)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
Given, \(\mathrm{T}=3.14=\frac{2 \pi}{\omega}\) \(\omega=2 \mathrm{rad} / \mathrm{s}\) \(\mathrm{x}=10 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=10 \omega \cos \left(\omega \mathrm{t}+\frac{\pi}{3}\right)\)…
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