Question

Two waves are represented by, $y_1=A\sin \left(kx-\omega t\right)$ and $y_2=A\cos \left(kx-\omega t\right)$. The amplitude of the resultant wave is:

• A. $4A$
• B. $2A$
• C. $\sqrt{2}A$
• D. $A$

Answer 1

The correct option is C

$\sqrt{2}A$

Step 1: Given

Equation of wave 1: $y_1=A\sin \left(kx-\omega t\right)$

Equation of wave 2: $y_2=A\cos \left(kx-\omega t\right)$

Amplitude of wave 1: $A_1=A$

Amplitude of wave 2: $A_2=A$

Step 2: Formula Used

Resultant Amplitude: $R=\sqrt{{A_1}^2+{A_2}^2+2A_1{A}_2\cos \varphi }$

Step 3: Find the phase difference between the waves by converting second equation into sine equation

$$\begin{gathered} y_2=A\cos \left(kx-\omega t\right) \\ =A\sin \left(kx-\omega t+\frac{\mathrm{\pi }}{2}\right) \\ =A\sin \left(kx-\omega t+\varphi \right) \\ \Rightarrow \varphi =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Step 4: Calculate the resultant amplitude using the formula

$$\begin{gathered} R=\sqrt{{A_1}^2+{A_2}^2+2A_1{A}_2\cos \varphi } \\ =\sqrt{A^2+A^2+2AA\cos \frac{\mathrm{\pi }}{2}} \\ =\sqrt{2A^2+0} \\ =\sqrt{2}A \end{gathered}$$

Hence, the amplitude of the resultant wave is $\sqrt{2}A$. Option C is correct.