Question

A vibratory motion is represented by

x = 2Acos wt + A cos $\left( \omega t + \frac{\pi}{2} \right)$+ A cos ($\omega$t + $\pi$)

+$\frac{A}{2}\cos\left( \omega t + \frac{3\pi}{2} \right)$

The resultant amplitude of the motion is

• A. $\frac{9A}{2}$
• B. $\frac{\sqrt{5}A}{2}$
• C. $\frac{5A}{2}$
• D. 2A

Answer 1

Answer: (B)

$\frac{\sqrt{5}A}{2}$

Answer 2

$x = 2A\cos\omega t + A\cos\left( \omega t + \frac{\pi}{2} \right) + A\cos(\omega t + \pi)$ $$+ \frac{A}{2}\cos(\omega t + \frac{3\pi}{2})$$

$$= 2A\cos\omega t - A\sin\omega t - A\cos\omega t + \frac{A}{2}\sin\omega t$$

$$= A\cos\omega t - \frac{A}{2}\sin\omega t$$

$\therefore$ the amplitude of the resultant motion is

$A_{R} = \sqrt{(A)^{2} + \left( - \frac{A}{2} \right)^{2}} = \frac{\sqrt{5}A}{2}$