Question

Three harmonic waves of same frequency (v) and intensity (I₀) having initial phase angles$0,\frac{\pi }{4},\frac{-\pi }{4}rad$ respectively. When they are superimposed, the resultant intensity is close to;

• A. $5.8I_0$
• B. $I_0$
• C. $3I_0$
• D. $0.2I_0$

Answer 1

The correct option is A

$5.8I_0$

Step 1: Given

Three harmonic waves of the same frequency (v) and intensity (I₀) having initial phase angles$0,\frac{\pi }{4},\frac{-\pi }{4}rad$

Step 2: Draw a diagram and apply the formula

As we know: $\text{Intensity}\left(I\right)\propto {\left(\text{Amplitude}\right)}^2$

$\begin{array}{l}I=K{A}^2\\ A^2=\frac{I}{K}\\ \therefore A=\frac{I}{K}\end{array}$

Where A is amplitude, I is the intensity, and K is constant.

Step 2: According to the question,

Refer above image,

$\begin{array}{c}{A}_{\text{resultant}}=(A_{1x}+A_{2x}+A_{3x})\widehat{i}+(A_{1y}+A_{2y}+A_{3y})\widehat{j}\\ =(A\cos \frac{\pi }{4}+A+A\cos \frac{\pi }{4})\widehat{i}+(A\sin \frac{\pi }{4}+0-A\sin \frac{\pi }{4})\widehat{j}\\ =(A+2A\cos 45°)\widehat{i}\\ =A\widehat{i}(1+\sqrt{2})\end{array}$

Amplitudes can be added using vector addition,

$A_{resul\tan t}=(\sqrt{2}+1)A$

So, for resultant

$\begin{array}{l}\therefore \frac{\sqrt{I_R}}{\sqrt{K}}=(1+\sqrt{2})\frac{\sqrt{I_0}}{\sqrt{K}}\\ \Rightarrow \sqrt{I_R}=(1+\sqrt{2})\sqrt{I_0}\\ \Rightarrow I_R=5.8I_0\end{array}$

Hence, the correct option is (A).