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Waves question

2023 · 24 Jan · Shift 1 · Q57
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  5. /2023 · 24 Jan · Shift 1 · Q57

Waves question

2023 · 24 Jan · Shift 1 · Q57

JEE MainPhysicsWavesMCQ+4 / −1
A travelling wave is described by the equation y(x,t)=[0.05sin⁡(8x−4t)]y(x,t) = [0.05\sin (8x - 4t)]y(x,t)=[0.05sin(8x−4t)] m The velocity of the wave is : [all the quantities are in SI unit]
  1. A
    4 ms−1\mathrm{4~ms^{-1}}4 ms−1
  2. B
    2 ms−1\mathrm{2~ms^{-1}}2 ms−1
  3. C
    8 ms−1\mathrm{8~ms^{-1}}8 ms−1
  4. D
    0.5 ms−1\mathrm{0.5~ms^{-1}}0.5 ms−1
View written solutionFree

Correct answer: D

  1. The standard form of a travelling wave is y(x,t)=Asin⁡(kx−ωt)y(x,t)=A\sin(kx-\omega t)y(x,t)=Asin(kx−ωt) where:

    • kkk = wave number
    • ω\omegaω = angular frequency
    • wave speed v=ωkv=\dfrac{\omega}{k}v=kω​

  2. Given: y(x,t)=0.05sin⁡(8x−4t)y(x,t)=0.05\sin(8x-4t)y(x,t)=0.05sin(8x−4t) Comparing with Asin⁡(kx−ωt)A\sin(kx-\omega t)Asin(kx−ωt), we get: k=8 rad m−1,ω=4 rad s−1k=8\ \text{rad m}^{-1},\qquad \omega=4\ \text{rad s}^{-1}k=8 rad m−1,ω=4 rad s−1

  3. Therefore, the wave speed is v=ωk=48=0.5 m s−1v=\frac{\omega}{k}=\frac{4}{8}=0.5\ \text{m s}^{-1}v=kω​=84​=0.5 m s−1

  4. Hence, the correct option is: D: 0.5 m s−1\boxed{\text{D: }0.5\ \text{m s}^{-1}}D: 0.5 m s−1​

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