Question

Three harmonic waves having equal frequency $v$ and same intensity $I_0$, have phase angles $0,\frac{\pi }{4}$ and $-\frac{\pi }{4}$ respectively. When they are superimposed the intensity of the resultant wave is close to

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

Answer 1

The correct option is B $5.8I_0$


As given phase angle of the three waves are $0,\frac{\pi }{4}$ and $-\frac{\pi }{4}$

Let the amplitude of each wave be $A$.

Resultant wave equation

$=A\text{sin}\omega t+A\text{sin}(\omega t-\frac{\pi }{4})+A\text{sin}(\omega t+\frac{\pi }{4})$

$=A\text{sin}\omega t+A\text{sin}\left(\omega t\right)\text{cos}\frac{\pi }{4}-A\text{cos}\left(\omega t\right)\text{sin}\frac{\pi }{4}+A\text{sin}\left(\omega t\right)\text{cos}\frac{\pi }{4}+A\text{cos}\left(\omega t\right)\text{sin}\frac{\pi }{4}$

$=A\text{sin}\omega t+\sqrt{2}A\text{sin}\omega t$

$=(\sqrt{2}+1)A\text{sin}\omega t$

As given, intensity of each wave is $I_0$

Resultant wave amplitude $=(\sqrt{2}+1)A$

As we know, $I\propto A^2$

So, $\frac{I}{I_0}=(\sqrt{2}+1{)}^2$

$I=5.8I_0$

Final answer : (a)