Question

The displacement x (in metre) of a particle in, simple harmonic motion is related to time t (in seconds) as $x = 0.01\cos\left( \pi t + \frac{\pi}{4} \right)$ The frequency of the motion will be

• A. 0.5 Hz
• B. 1.0 Hz
• C. $\frac{\pi}{2}Hz$
• D. $\pi Hz$

Answer 1

Answer: (A)

0.5 Hz

Answer 2

Comparing given equation with standard equation,$x = a\cos(\omega t + \varphi)$ we get, $a = 0.01$ and$\omega = \pi$

$\Rightarrow 2\pi n = \pi \Rightarrow n = 0.5Hz$