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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2019
  • Medium
  • MCQ

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

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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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