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Waves question

2022 · 28 Jul · Shift 1 · Q54
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Waves question

2022 · 28 Jul · Shift 1 · Q54

JEE MainPhysicsWavesMCQ+4 / −1
In the wave equation y=0.5sin⁡2πλ(400t−x) my=0.5 \sin \frac{2 \pi}{\lambda}(400 \mathrm{t}-x) \,\mathrm{m}y=0.5sinλ2π​(400t−x)m the velocity of the wave will be:
  1. A
    200 m/s
  2. B
    200 2\sqrt22​ m/s
  3. C
    400 m/s
  4. D
    400 2\sqrt22​ m/s
View written solutionFree

Correct answer: C

  1. Write the given wave equation in standard form

The given wave is

y=0.5sin⁡(2πλ(400t−x))y = 0.5\sin\left(\frac{2\pi}{\lambda}(400t - x)\right)y=0.5sin(λ2π​(400t−x))

We compare it with the standard progressive wave form:

y=Asin⁡(2πλ(vt−x))y = A\sin\left(\frac{2\pi}{\lambda}(vt - x)\right)y=Asin(λ2π​(vt−x))

where vvv is the wave speed.

  1. Identify the wave speed

From the comparison,

v=400 m/sv = 400\ \text{m/s}v=400 m/s
  1. Alternative verification using } \omega \text{ and } k

The standard form can also be written as

y=Asin⁡(ωt−kx)y = A\sin(\omega t - kx)y=Asin(ωt−kx)

with

k=2πλ,ω=2πλ⋅400k = \frac{2\pi}{\lambda}, \qquad \omega = \frac{2\pi}{\lambda}\cdot 400k=λ2π​,ω=λ2π​⋅400

So,

v=ωk=2πλ⋅4002πλ=400 m/sv = \frac{\omega}{k} = \frac{\frac{2\pi}{\lambda}\cdot 400}{\frac{2\pi}{\lambda}} = 400\ \text{m/s}v=kω​=λ2π​λ2π​⋅400​=400 m/s
  1. Check the options
  • A: 200 m/s200\ \text{m/s}200 m/s ❌
  • B: 2002 m/s200\sqrt{2}\ \text{m/s}2002​ m/s ❌
  • C: 400 m/s400\ \text{m/s}400 m/s ✅
  • D: 4002 m/s400\sqrt{2}\ \text{m/s}4002​ m/s ❌

Therefore, the correct option is C.

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