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JEE Main 2021 (26 Aug, Shift 2) · Reflection, Superposition & Interference

Reflection, Superposition & Interference question from JEE Main 2021 (26 Aug, Shift 2)

Q16JEE Main 2021 · 26 AUG · SHIFT 2Medium

Reflection, Superposition & Interference

Two waves are simultaneously passing through a string and their equations are: y1 = A1 sin k(x $-$ vt), y2 = A2 sin k(x $-$ vt + x0). Given amplitudes A1 = 12 mm and A2 = 5 mm, x0 = 3.5 cm and wave number k = 6.28 cm ${}^{-1}$ . The amplitude of resulting wave will be ................ mm.

Numerical answer type. Work it out, then open the answer.

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Answer: 7

Phase difference $\phi = kx_0 = 6.28 \times 3.5 = 2\pi \times 3.5 = 7\pi$ , i.e. opposite phase.
$A_R = A_1 - A_2 = 12 - 5 = 7\text{ mm}$ .

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