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Waves question

2023 · 25 Jan · Shift 1 · Q68
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Waves question

2023 · 25 Jan · Shift 1 · Q68

JEE MainPhysicsWavesNumerical+4 / −1
The distance between two consecutive points with phase difference of 60 ∘^\circ∘ in a wave of frequency 500 Hz is 6.0 m. The velocity with which wave is travelling is ‾\underline{\hspace{2cm}}​ km/s
Numerical answer
View written solutionFree

Correct answer: 18

  1. For a progressive wave, phase difference Δϕ\Delta \phiΔϕ and path difference Δx\Delta xΔx are related by
Δϕ=2πλΔx\Delta \phi = \frac{2\pi}{\lambda} \Delta xΔϕ=λ2π​Δx
  1. Here, the phase difference is 60∘=π360^\circ = \frac{\pi}{3}60∘=3π​ and the distance between the two points is Δx=6.0 m\Delta x = 6.0\,\text{m}Δx=6.0m.

So,

π3=2πλ(6)\frac{\pi}{3} = \frac{2\pi}{\lambda}(6)3π​=λ2π​(6)
  1. Solve for wavelength λ\lambdaλ:
13=12λ\frac{1}{3} = \frac{12}{\lambda}31​=λ12​ λ=36 m\lambda = 36\,\text{m}λ=36m
  1. Wave speed is given by
v=fλv = f\lambdav=fλ

Given f=500 Hzf = 500\,\text{Hz}f=500Hz,

v=500×36=18000 m/sv = 500 \times 36 = 18000\,\text{m/s}v=500×36=18000m/s
  1. Convert to km/s:
18000 m/s=18 km/s18000\,\text{m/s} = 18\,\text{km/s}18000m/s=18km/s

Therefore, the wave velocity is

18\boxed{18}18​

km/s.

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