Question

The amplitude of a wave represented by displacement equation $y=\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \cos \omega t$ will be

• A. $\frac{a+b}{a b}$
• B. $\frac{\sqrt{a}+\sqrt{b}}{a b}$
• C. $\frac{\sqrt{a} \pm \sqrt{b}}{a b}$
• D. $\sqrt{\frac{a+b}{a b}}$

Answer 1

Answer: (D)

$y=\frac{1}{\sqrt{a}} \sin \omega t \pm \frac{1}{\sqrt{b}} \sin \left(\omega t+\frac{\pi}{2}\right)$
Here phase difference $=\frac{\pi}{2}$
Resultant amp. $=\sqrt{\left(\frac{1}{\sqrt{a}}\right)^{2}+\left(\frac{1}{\sqrt{b}}\right)^{2}}$
$=\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a+b}{a b}}$