JEE Main On 16 April 2018 Question 21
Question: A particle executes simple harmonic motion and is located at $ x=a,b $ and c at times $ t _0,2t _0 $ and $ 3t _0 $ respectively. The frequency of the oscillation is [JEE Main 16-4-2018]
Options:
A) $ \frac{1}{2\pi t _0}{{\cos }^{-1}}( \frac{a+b}{2c} ) $
B) $ \frac{1}{2\pi t _0}{{\cos }^{-1}}( \frac{a+b}{3c} ) $
C) $ \frac{1}{2\pi t _0}{{\cos }^{-1}}( \frac{2a+3c}{b} ) $
D) $ \frac{1}{2\pi t _0}{{\cos }^{-1}}( \frac{a+c}{2b} ) $
Show Answer
Answer:
Correct Answer: D
Solution:
$ a=A\cos \omega t _{o} $ …………(1)
$ b=A\cos 2\omega t _{o} $ ………….(2)
$ c=A\cos 3\omega t _{o} $ ………….(3)
On adding (1) and (3) $ a+c=A(\cos \omega t _{o}+\cos 3\omega t _{o}) $
$ a+c=2A\cos (\frac{3\omega t _{o}+\omega t _{o}}{2})cos(\frac{3\omega t _{o}-\omega t _{o}}{2}) $
$ a+c=2A\cos 2\omega t _{o}\cos \omega t _{o} $ from (2), $ b=A\cos 2\omega t _{o} $
$ a+c=2b\cos \omega t _{o} $ $ {{\cos }^{-1}}(\frac{a+c}{2b})=2\pi ft _{o} $
$ f=\frac{1}{2\pi t _{o}}{{\cos }^{-1}}(\frac{a+c}{2b}) $
