JEE Main 2025 Apr 8 Shift 2, Physics Q17: Superposition of Waves

JEE Main2025Apr 8, Shift 2Physics

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, $y_{1} (x, t) = 4 sin (k x - ω t)$ and $y_{2} (x, t) = 2 sin (k x - ω t + \frac{2 π}{3})$ , are:

(Take the angular frequency of initial waves same as $ω$ )

A
$[6, \frac{2 π}{3}]$
B
$[6, \frac{π}{3}]$
C
$[3, \frac{π}{6}]$
D
$[23, \frac{π}{6}]$

Solution

Represent each wave as a phasor: $y_{1}$ has amplitude 4 at phase 0 and $y_{2}$ has amplitude 2 at phase $2 π /3$ ; the angle between them is $12 0^{\circ}$ .

Resultant amplitude (law of cosines): $A = 4^{2} + 2^{2} + 2 \cdot 4 \cdot 2 cos 12 0^{\circ} = 16 + 4 - 8 = 12 = 23$ .

Resultant phase relative to $y_{1}$ : $tan ϕ = \frac{2 sin 12 0 ^{\circ}}{4 + 2 cos 12 0 ^{\circ}} = \frac{3}{3} = \frac{1}{3}$ .

So $ϕ = \frac{π}{6}$ .

The resultant has amplitude $23$ and phase $\frac{π}{6}$ , matching option (4).

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JEE Main 2025 Apr 8 Shift 2, Physics Q17: Superposition of Waves

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