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)

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