JEE Mains · Physics · STD 11 - 13. oscillations
A simple harmonic motion is represented by \(y\, = 5\,(\sin \,3\pi t\, + \,\sqrt 3 \,\cos \,3\pi t)\,cm\) The amplitude and time period of the motion are
- A \(10\,\,cm\,,\,\frac{2}{3}\,s\)
- B \(10\,\,cm\,,\,\frac{3}{2}\,s\)
- C \(5\,\,cm\,,\,\frac{3}{2}\,s\)
- D \(5\,\,cm\,,\,\frac{2}{3}\,s\)
Answer & Solution
Correct Answer
(A) \(10\,\,cm\,,\,\frac{2}{3}\,s\)
Step-by-step Solution
Detailed explanation
\(y=5(\sin 3 \pi t+\sqrt{3} \cos 3 \pi t) \mathrm{cm}\) \(\Rightarrow \quad y=10 \sin (3 \pi t+\phi)\) \(\Rightarrow A=10 \mathrm{cm}\) \(\Rightarrow \quad T=\frac{2}{3} \sec\)
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