| A simple harmonic motion is represented by [JEE Main 12-Jan-2019 Evening] |
| \[y=5(sin3\pi t+\sqrt{3}cos3\pi t)cm\] |
| The amplitude and time period of the motion are |
A) \[5cm,\frac{3}{2}s\]
B) \[10cm,\frac{2}{3}s\]
C) \[5cm,\frac{2}{3}s\]
D) \[10cm,\frac{3}{2}s\]
Correct Answer: B
Solution :
| [b] The given equation is y = 5 |
| \[(sin3\pi t+\sqrt{3}cos3\pi t)\] |
| \[\Rightarrow \]\[y=2\times 5\left( \frac{1}{2}\sin 3\pi t+\frac{\sqrt{3}}{2}\cos 3\pi t \right)\] |
| \[=10\left( \cos \frac{\pi }{3}\sin 3\pi t+\sin \frac{\pi }{3}\cos 3\pi t \right)\] |
| \[y=10\sin \left( 3\pi t+\frac{\pi }{3} \right)\] .(i) |
| Comparing eqn. (i) with standard equation |
| \[y=A\sin (\omega t+\phi )\] |
| \[\Rightarrow \]\[\omega =3\pi \]and \[A=10cm\] |
| \[\therefore \]Time period \[T=\frac{2\pi }{\omega }=\frac{2}{3}s\] |
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