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

Answer 1

The correct option is B $\sqrt{\frac{a+b}{ab}}$

Given:
$y=\frac{1}{\sqrt{a}}\sin \omega t\pm \frac{1}{\sqrt{b}}\cos \omega t\dots \left(i\right)$
Equation $\left(i\right)$ can be written as,
$y=\frac{1}{\sqrt{a}}\sin \omega t\pm \frac{1}{\sqrt{b}}\sin (\omega t+\frac{\pi }{2})$
Hence, phase difference $=\frac{\pi }{2}$
Therefore, the resultant amplitude =

$\sqrt{{\left(\frac{1}{\sqrt{a}}\right)}^2}+\sqrt{{\left(\frac{1}{\sqrt{b}}\right)}^2}=\sqrt{\frac{1}{a}+\frac{1}{b}}$

$=\sqrt{\frac{a+b}{ab}}$

Final answer: $\left(d\right)$