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Waves question

2025 · 8 Apr · Shift 2 · Q58
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Waves question

2025 · 8 Apr · Shift 2 · Q58

JEE MainPhysicsWavesMCQ+4 / −1
The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves, y1(x,t)=4sin⁡(kx−ωt)y_1(x, t) = 4 \sin (kx - \omega t)y1​(x,t)=4sin(kx−ωt) and y2(x,t)=2sin⁡(kx−ωt+2π3)y_2(x, t) = 2 \sin (kx - \omega t + \frac{2\pi}{3})y2​(x,t)=2sin(kx−ωt+32π​), are: (Take the angular frequency of initial waves same as ω\omegaω)
  1. A
    [3,π6]\left[\sqrt{3}, \frac{\pi}{6}\right][3​,6π​]
  2. B
    [23,π6]\left[2\sqrt{3}, \frac{\pi}{6}\right][23​,6π​]
  3. C
    [6,2π3]\left[6, \frac{2\pi}{3}\right][6,32π​]
  4. D
    [6,π3]\left[6, \frac{\pi}{3}\right][6,3π​]
View written solutionFree

Correct answer: B

  1. Write the two waves in a common form

Let

θ=kx−ωt.\theta = kx - \omega t.θ=kx−ωt.

Then the given waves are

y1=4sin⁡θ,y_1 = 4\sin\theta,y1​=4sinθ, y2=2sin⁡(θ+2π3).y_2 = 2\sin\left(\theta + \frac{2\pi}{3}\right).y2​=2sin(θ+32π​).

We want the resultant in the form

y=Rsin⁡(θ+ϕ),y = R\sin(\theta + \phi),y=Rsin(θ+ϕ),

where RRR is the amplitude and ϕ\phiϕ is the phase.

  1. Expand the second wave

Using

sin⁡(θ+α)=sin⁡θcos⁡α+cos⁡θsin⁡α,\sin(\theta+\alpha)=\sin\theta\cos\alpha+\cos\theta\sin\alpha,sin(θ+α)=sinθcosα+cosθsinα,

with α=2π3\alpha=\frac{2\pi}{3}α=32π​,

cos⁡2π3=−12,sin⁡2π3=32.\cos\frac{2\pi}{3}=-\frac{1}{2}, \qquad \sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2}.cos32π​=−21​,sin32π​=23​​.

So,

y2=2(sin⁡θcos⁡2π3+cos⁡θsin⁡2π3)y_2 = 2\left(\sin\theta\cos\frac{2\pi}{3}+\cos\theta\sin\frac{2\pi}{3}\right)y2​=2(sinθcos32π​+cosθsin32π​) =2(−12sin⁡θ+32cos⁡θ)=2\left(-\frac{1}{2}\sin\theta+\frac{\sqrt{3}}{2}\cos\theta\right)=2(−21​sinθ+23​​cosθ) =−sin⁡θ+3cos⁡θ.=-\sin\theta+\sqrt{3}\cos\theta.=−sinθ+3​cosθ.

Hence,

y=y1+y2=4sin⁡θ+(−sin⁡θ+3cos⁡θ)y = y_1+y_2 = 4\sin\theta + \left(-\sin\theta+\sqrt{3}\cos\theta\right)y=y1​+y2​=4sinθ+(−sinθ+3​cosθ) =3sin⁡θ+3cos⁡θ.=3\sin\theta+\sqrt{3}\cos\theta.=3sinθ+3​cosθ.
  1. Compare with the standard form

If

y=Rsin⁡(θ+ϕ),y=R\sin(\theta+\phi),y=Rsin(θ+ϕ),

then

y=Rsin⁡θcos⁡ϕ+Rcos⁡θsin⁡ϕ.y=R\sin\theta\cos\phi + R\cos\theta\sin\phi.y=Rsinθcosϕ+Rcosθsinϕ.

Comparing with

y=3sin⁡θ+3cos⁡θ,y=3\sin\theta+\sqrt{3}\cos\theta,y=3sinθ+3​cosθ,

we get

Rcos⁡ϕ=3,R\cos\phi = 3,Rcosϕ=3, Rsin⁡ϕ=3.R\sin\phi = \sqrt{3}.Rsinϕ=3​.
  1. Find the amplitude

Square and add:

R2=(Rcos⁡ϕ)2+(Rsin⁡ϕ)2=32+(3)2=9+3=12.R^2 = (R\cos\phi)^2 + (R\sin\phi)^2 = 3^2 + (\sqrt{3})^2 = 9+3=12.R2=(Rcosϕ)2+(Rsinϕ)2=32+(3​)2=9+3=12.

So,

R=12=23.R=\sqrt{12}=2\sqrt{3}.R=12​=23​.
  1. Find the phase
tan⁡ϕ=Rsin⁡ϕRcos⁡ϕ=33=13.\tan\phi = \frac{R\sin\phi}{R\cos\phi} = \frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}.tanϕ=RcosϕRsinϕ​=33​​=3​1​.

Thus,

ϕ=π6.\phi = \frac{\pi}{6}.ϕ=6π​.
  1. Final result

Therefore the resultant wave is

y=23sin⁡(kx−ωt+π6).y = 2\sqrt{3}\sin\left(kx-\omega t + \frac{\pi}{6}\right).y=23​sin(kx−ωt+6π​).

So the amplitude is 232\sqrt{3}23​ and the phase is π6\frac{\pi}{6}6π​.

Hence the correct option is B.

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