The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, y₁(x, t) = 4sin(kx−ωt) and y₂(x, t) = 2sin(kx− ωt+ 2π/3), are:

 

🌊 Superposition of Two Harmonic Waves – Find Resultant Amplitude & Phase | JEE Physics | Doubtify

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📌 Question:

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves,
y₁(x, t) = 4sin(kx − ωt) and
y₂(x, t) = 2sin(kx − ωt + 2π/3) are:

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🖼️ Question Image:

🧠 Solution Image:

📖 Concept:

This is a classic question on the principle of superposition of waves.
When two harmonic waves of the same frequency and wave vector (k) interfere, but have different amplitudes and a phase difference, their resultant wave is also a harmonic wave.

We use phasor addition to find:

• Resultant Amplitude (A):

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

• Resultant Phase (Φ):

$$\tan \mathrm{\Phi }=\frac{A_2\sin \varphi }{A_1+A_2\cos \mathrm{\varphi }}$$

Where:

• A₁ = 4

• A₂ = 2

• ϕ = 2π/3

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🧮 Calculation:

Step 1:

$$A=\sqrt{4^2+2^2+2\times 4\times 2\times \cos (2\pi \mathrm{/}3)}=\sqrt{16+4+16\times (-1\mathrm{/}2)}=\sqrt{20-8}=\sqrt{12}=2\sqrt{\mathrm{3}}$$

Step 2:

$$\tan \mathrm{\Phi }=\frac{2\sin (2\pi \mathrm{/}3)}{4+2\cos (2\pi \mathrm{/}3)}=\frac{2\times \sqrt{3}\mathrm{/}2}{4-1}=\frac{\sqrt{3}}{3}⟹\mathrm{\Phi }={30}^{\circ }=\frac{\pi }{\mathrm{6}}$$

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✅ Final Answer:

Amplitude = 2√3, Phase = π/6

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🔍 Why This Question Is Important:

• Appears frequently in JEE Mains and Advanced

• Tests your conceptual grip on wave interference and phasor diagrams

• Helps in mastering wave superposition for future questions in sound and optics

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🧠 Pro Tip:

Whenever two waves have the same frequency and direction, use phasor addition (vector addition) for amplitude and phase — don't just add their equations blindly!

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