Animated Solution for Physics - Waves: Three harmonic waves having equal frequency $\nu$ 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

Imagine you are standing on a beach, and three distinct waves hit you at the exact same moment. How do you calculate the total impact?

When dealing with the superposition of multiple harmonic waves, adding their trigonometric equations algebraically can quickly become a nightmare. This is where the elegance of phasors comes in.

A phasor is simply a rotating vector. The length of this vector represents the amplitude of the wave, and the angle it makes with the horizontal axis represents its phase. By converting waves into vectors, we turn a complex calculus problem into simple geometry!

In our problem, we have three waves with the same frequency $ u$ and the same initial intensity $I_0$. Since intensity is proportional to the square of the amplitude ($I_0\propto A^2$), all three waves have the same amplitude $A$.

Let's draw their phasors on a coordinate system: 1. The first wave has a phase of $0$, so its phasor points directly along the positive X-axis. 2. The second wave has a phase of $\frac{\pi }{4}$, so its phasor points diagonally upwards at $45^{\circ }$. 3. The third wave has a phase of $-\frac{\pi }{4}$, so its phasor points diagonally downwards at $-45^{\circ }$.

Look closely at the two angled phasors. Because they are at $\frac{\pi }{4}$ and $-\frac{\pi }{4}$, their vertical components are exactly equal and opposite:

They perfectly cancel each other out! We only need to add their horizontal components:

Now, we have a combined phasor of length $\sqrt{2}A$ pointing right along the X-axis.

But wait, our first wave's phasor is also sitting right there on the X-axis! Since they point in the exact same direction, we can just add their lengths directly to find the total resultant amplitude:

The question asks for the resultant intensity, not just the amplitude. We know that intensity is directly proportional to the square of the amplitude ($I\propto A^2$).

So, we square our total amplitude:

Expanding this using the $(a+b{)}^2$ identity:

Now, let's substitute the approximate value of $\sqrt{2}\approx 1.414$:

Rounding off to one decimal place, we get our final answer: $I_R\approx 5.8I_0$

This problem beautifully demonstrates how visualizing physics through geometry can make even the most daunting calculations incredibly simple!