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Waves question

2008 · Shift 0 · Q74
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Waves question

2008 · Shift 0 · Q74

JEE MainPhysicsWavesMCQ+4 / −1
A wave travelling along the xxx-axis is described by the equation y(x,t)=0.005y(x, t)=0.005y(x,t)=0.005 cos⁡ (α x−βt).\cos \,\left( {\alpha \,x - \beta t} \right).cos(αx−βt). If the wavelength and the time period of the wave are 0.08m0.08m0.08m and 2.0s2.0s2.0s, respectively, then α\alphaα and β\betaβ in appropriate units are
  1. A
    α=25.00π, β=π\alpha = 25.00\pi ,\,\beta = \piα=25.00π,β=π
  2. B
    α=0.08π, β=2.0π\alpha = {{0.08} \over \pi },\,\beta = {{2.0} \over \pi }α=π0.08​,β=π2.0​
  3. C
    α=0.04π, β=1.0π\alpha = {{0.04} \over \pi },\,\beta = {{1.0} \over \pi }α=π0.04​,β=π1.0​
  4. D
    α=12.50π, β=π2.0\alpha = 12.50\pi ,\,\beta = {\pi \over {2.0}}α=12.50π,β=2.0π​
View written solutionFree

Correct answer: A

  1. Compare with the standard wave equation

A travelling wave along the xxx-axis is generally written as

y(x,t)=Acos⁡(kx−ωt),y(x,t)=A\cos(kx-\omega t),y(x,t)=Acos(kx−ωt),

where:

  • kkk is the wave number,
  • ω\omegaω is the angular frequency.

Given:

y(x,t)=0.005cos⁡(αx−βt).y(x,t)=0.005\cos(\alpha x-\beta t).y(x,t)=0.005cos(αx−βt).

So, by comparison,

α=k,β=ω.\alpha = k, \qquad \beta = \omega.α=k,β=ω.
  1. Use the relation for wavelength

Wave number is

k=2πλ.k=\frac{2\pi}{\lambda}.k=λ2π​.

Given

λ=0.08 m.\lambda=0.08\,\text{m}.λ=0.08m.

Therefore,

α=k=2π0.08=25π.\alpha = k = \frac{2\pi}{0.08} = 25\pi.α=k=0.082π​=25π.
  1. Use the relation for time period

Angular frequency is

ω=2πT.\omega=\frac{2\pi}{T}.ω=T2π​.

Given

T=2.0 s.T=2.0\,\text{s}.T=2.0s.

Therefore,

β=ω=2π2.0=π.\beta = \omega = \frac{2\pi}{2.0} = \pi.β=ω=2.02π​=π.
  1. Match with the options

Thus,

α=25.00π,β=π.\alpha = 25.00\pi, \qquad \beta = \pi.α=25.00π,β=π.

This matches Option A.

  1. Verification with stored answer

Stored correct answer: A

Our derived answer: A

So, the derived answer agrees with the stored correct answer.

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