Question

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

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

Answer 1

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

Given ,
$y=\frac{1}{\sqrt{a}}\sin (kx-\omega t)\pm \frac{1}{\sqrt{b}}\cos (kx-\omega t)$
We can re write the above equation as,
$y=\frac{1}{\sqrt{a}}\sin (kx-\omega t)\pm \frac{1}{\sqrt{b}}\sin (\frac{\pi }{2}+kx-\omega t)$
This is a superposition of two waves, having phase difference $\varphi =\frac{\pi }{2}$

$\therefore$ Resultant amplitude
$R=\sqrt{{\left(A\right)}^2+{\left(B\right)}^2+2AB\cos \varphi }$

$\Rightarrow R=\sqrt{{\left(\frac{1}{\sqrt{a}}\right)}^2+{\left(\frac{1}{\sqrt{b}}\right)}^2+2\times \frac{1}{\sqrt{a}}\times \frac{1}{\sqrt{b}}\cos {90}^{\circ }}$
$\Rightarrow R=\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a+b}{ab}}$
Thus, option (d) is the correct answer.