Resultant Amplitude and Phase of Two Waves

❓ Question

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves:

$$y_1=4\sin(kx-\omega t)$$

and

$$y_2=2\sin(kx-\omega t+\pi/2)$$

are:

(Take angular frequency of initial waves same as $\omega$)

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Superposition principle
👉 Resultant amplitude
👉 Phase difference of SHM.

Main idea:

For two waves:

$$A_1\sin\theta$$

and

$$A_2\sin(\theta+\phi)$$

Resultant amplitude:

$$A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$$

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🔷 Step 1 — Identify Given Values 💯

Given:

$$A_1=4$$
$$A_2=2$$

Phase difference:

$$\phi=\frac{\pi}{2}$$

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🔷 Step 2 — Calculate Resultant Amplitude

Using formula:

$$A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi}$$

Substitute values:

$$A= \sqrt{ 4^2+2^2+2(4)(2)\cos\frac{\pi}{2} }$$

Since:

$$\cos\frac{\pi}{2}=0$$

So:

$$A=\sqrt{16+4}$$
$$A=\sqrt{20}$$
$$A=2\sqrt5$$

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🔷 Step 3 — Find Resultant Phase

Resultant phase angle:

$$\tan\theta= \frac{A_2\sin\phi}{A_1+A_2\cos\phi}$$

Substitute values:

$$\tan\theta= \frac{2\sin(\pi/2)}{4+2\cos(\pi/2)}$$
$$= \frac{2}{4}$$ $$=\frac12$$

Thus:

$$\theta=\tan^{-1}\left(\frac12\right)$$

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🔷 Step 4 — Final Wave

Resultant wave:

$$y=2\sqrt5\sin\left(kx-\omega t+\tan^{-1}\frac12\right)$$

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🔷 Step 5 — Correct Option

$$\boxed{ \left(2\sqrt5,\ \tan^{-1}\frac12\right) }$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Direct amplitudes add kar dena

❌ Phase difference ignore kar dena

❌ $\cos\frac{\pi}{2}=1$ mistake kar dena

Remember:

$$\boxed{{\displaystyle A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }}}$$

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✅ Final Answer

$$\boxed{ 2\sqrt5,\ \tan^{-1}\left(\frac12\right) }$$

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📚 Related Topics

• Resultant Amplitude with Phase Difference π/2
• Interference Pattern with Red and Green Light
• Interference of Light Waves Using Phase Difference