Q195 / 200 · Manipal (MET) 2012 · Physics
Two waves are represented by $y_1=A \sin (k x-\omega t)$ and $y_2=A \cos (k x-\omega t)$. The amplitude of resultant wave is
- A$4 A$
- B$2 A$
- Correct answerC$\sqrt{2} A$
- D$A$
Solution
Given,
$\begin{aligned}
& y_1=A \sin (k x-\omega t) \\
& y_2=A \cos (k x-\omega t)
\end{aligned}$
$\text { or } \quad y_2=A \sin \left(k x-\omega t+\frac{\pi}{2}\right)$
Phase difference of two waves $=\frac{\pi}{2}$ $\because \quad$ Resultant amplitude $\begin{aligned} R & =\sqrt{A^2+A^2+2 A A \cos \phi} \\ & =\sqrt{A^2+A^2+2 A^2 \cos \frac{\pi}{2}} \\ & =\sqrt{2 A^2} \quad\left(\because \cos \frac{\pi}{2}=0\right) \\ R & =\sqrt{2}A \end{aligned}$
Phase difference of two waves $=\frac{\pi}{2}$ $\because \quad$ Resultant amplitude $\begin{aligned} R & =\sqrt{A^2+A^2+2 A A \cos \phi} \\ & =\sqrt{A^2+A^2+2 A^2 \cos \frac{\pi}{2}} \\ & =\sqrt{2 A^2} \quad\left(\because \cos \frac{\pi}{2}=0\right) \\ R & =\sqrt{2}A \end{aligned}$
Hence, the correct answer is option: 'C'.