Question

The simple harmonic motion of a particle is represented by the equation x = 4 cos $ \left[ 88t+\frac{\pi }{4} \right]. $ The frequency (in Hz) and the initial displacement (in m) of the particle are:

• A. 14, 2 $ \sqrt{2} $
• B. 16, 2 $ \sqrt{2} $
• C. 14, 3 $ \sqrt{2} $
• D. 16, 3 $ \sqrt{2} $

Answer 1

Answer: (A)

$ x=4\cos \left[ 88\,t+\frac{\pi }{4} \right] $ ?(i) Comparing Eq. (i) with standard equation of SHM of a particle, given by $ x=A\cos (\omega t+\text{o }\!\!|\!\!\text{ }) $ ...(ii) We have, $ \omega =88 $ $ \Rightarrow $ $ f=\frac{88}{2\pi }=14\,Hz. $ From Eq. (i), we have $ x=4\left[ \cos 88t\,\cos \frac{\pi }{4}-\sin 88t\sin \frac{\pi }{4} \right] $ $ =4\left[ \cos 88t\times \frac{1}{\sqrt{2}}-\sin 88t\times \frac{1}{\sqrt{2}} \right] $ So, initial displacement of the particle is $ 2\sqrt{2}\,m. $