Question

Two waves are represented by $y_1=A\sin t$ and $y_2=A\cos t$. The phase of the first wave, w.r.t. to the second wave is:

• A. More by $\mathrm{\pi }\mathrm{radian}$
• B. Less by $\mathrm{\pi }\mathrm{radian}$
• C. More by $\frac{\mathrm{\pi }}{2}$
• D. Less by $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is D

Less by $\frac{\mathrm{\pi }}{2}$

Step 1: Given

Wave equation of first wave: $y_1=A\sin t$

Wave equation of second wave: $y_2=A\cos t$

The phase of first wave is ${\varphi }_1$

The phase of second wave is ${\varphi }_2$

Step 2: Formula used

The wave equation is represented as $y\left(x,t\right)=A\sin \left(kx-\omega t+\varphi \right)$, where $y\left(x,t\right)$ is vertical position, $A$ is amplitude, $k$ is angular frequency number, $x$ is horizontal position variable, $\omega$ is angular frequency, $t$ is time variable and $\varphi$ is phase.

$\cos x=\sin \left(x-\frac{\mathrm{\pi }}{2}\right)$

Step 3: Find the phase difference

Convert the wave equation of second wave in sine form

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

Compare $y_1$ with wave equation and find phase.

$$\begin{gathered} y\left(x,t\right)=y_1 \\ \Rightarrow A\sin \left(kx-\omega t+\varphi \right)=A\sin t \\ \Rightarrow {\varphi }_1=0 \end{gathered}$$

Compare $y_2$ with wave equation and find phase.

$$\begin{gathered} y\left(x,t\right)=y_2 \\ \Rightarrow A\sin \left(kx-\omega t+\varphi \right)=A\sin \left(t-\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow {\varphi }_2=\frac{\mathrm{\pi }}{2} \end{gathered}$$

Calculate the phase difference by subtracting phase of first wave from that of second.

$$\begin{gathered} \varphi ={\varphi }_2-{\varphi }_1 \\ =\frac{\mathrm{\pi }}{2}-0 \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Hence, the phase difference is $\frac{\mathrm{\pi }}{2}$.