Question

The amplitude of a wave represented by the 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. $\frac{\sqrt{a}+\sqrt{b}}{ab}$
  
• 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 :
$y=\frac{1}{\sqrt{a}}\sin \omega t\pm \frac{1}{\sqrt{b}}\cos \omega t$ ...(i)

Let,
$\frac{1}{\sqrt{a}}=A\cos \theta$ ...(ii)

$\frac{1}{\sqrt{b}}=A\sin \theta$ ..(iii)

$\Rightarrow y=A\cos \theta \sin \omega t\pm A\sin \theta \cos \omega t$

$\therefore y=A\sin (\omega t\pm \theta )$

Squaring and adding Eqs. (ii) and (iii):
$A^2{\cos }^2\theta +A^2{\sin }^2\theta =\frac{1}{a}+\frac{1}{b}$
$\Rightarrow A=\sqrt{\frac{a+b}{ab}}$

Hence, $\left(D\right)$ is the correct answer.