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NVAEasyJEE 2023Superposition Principle & Standing Waves

JEE Physics 2023 Question with Solution

Two simple harmonic waves having equal amplitudes of 8 cm8 \, \text{cm} and equal frequency of 10 Hz10 \, \text{Hz} are moving along the same direction. The resultant amplitude is also 8 cm8 \, \text{cm}. The phase difference between the individual waves is —– degrees.

Answer

Correct answer:120

Step-by-step solution

Standard Method

Given: Two simple harmonic waves have amplitudes A1=A2=8 cmA_1 = A_2 = 8 \, \text{cm}, frequency 10 Hz10 \, \text{Hz}, and resultant amplitude Aresultant=8 cmA_{\text{resultant}} = 8 \, \text{cm}.

Find: The phase difference ϕ\phi between the two waves.

For the resultant amplitude of two waves,

Aresultant=A12+A22+2A1A2cos⁡ϕA_{\text{resultant}} = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos \phi}

Substitute the given values:

8=82+82+2×8×8cos⁡ϕ8 = \sqrt{8^2 + 8^2 + 2 \times 8 \times 8 \cos \phi}

Squaring both sides,

64=64+64+128cos⁡ϕ64 = 64 + 64 + 128 \cos \phi 64=128+128cos⁡ϕ64 = 128 + 128 \cos \phi 128cos⁡ϕ=−64128 \cos \phi = -64 cos⁡ϕ=−12\cos \phi = -\frac{1}{2}

Therefore,

ϕ=120∘\phi = 120^\circ

So, the phase difference between the individual waves is 120∘120^\circ.

Common mistakes

  • Using simple addition of amplitudes, A=A1+A2A = A_1 + A_2, is wrong because the waves have a phase difference. Use the resultant amplitude formula involving cos⁡ϕ\cos \phi instead.

  • Forgetting to square both sides after substituting into the square root expression leads to an incorrect equation. First remove the square root correctly, then solve for cos⁡ϕ\cos \phi.

  • Taking cos⁡ϕ=−12\cos \phi = -\frac{1}{2} and choosing the wrong angle is incorrect. In the standard range for phase difference, ϕ=120∘\phi = 120^\circ is the appropriate value, not 60∘60^\circ.

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