JEE Main 12 Jan 2019 Evening Question 4
Question: A simple harmonic motion is represented by $ y=5(sin3\pi t+\sqrt{3}cos3\pi t)cm $ .The amplitude and time period of the motion are
[JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) $ 5cm,\frac{3}{2}s $
B) $ 10cm,\frac{2}{3}s $
C) $ 5cm,\frac{2}{3}s $
D) $ 10cm,\frac{3}{2}s $
Show Answer
Answer:
Correct Answer: B
Solution:
-
The given equation is y = 5 $ (sin3\pi t+\sqrt{3}cos3\pi t) $
$ \Rightarrow $ $ y=2\times 5( \frac{1}{2}\sin 3\pi t+\frac{\sqrt{3}}{2}\cos 3\pi t ) $ $ =10( \cos \frac{\pi }{3}\sin 3\pi t+\sin \frac{\pi }{3}\cos 3\pi t ) $
$ y=10\sin ( 3\pi t+\frac{\pi }{3} ) $ -.(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 $
