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JEE Physics 2023 Question with Solution

A travelling wave is described by the equation: y(x,t)=0.05sin⁡(8x−4t) my(x, t) = 0.05 \sin(8x - 4t) \, \text{m}. The velocity of the wave is: (All quantities are in SI units.)

  • A

    4 ms−14 \, \text{ms}^{-1}

  • B

    2 ms−12 \, \text{ms}^{-1}

  • C

    0.5 ms−10.5 \, \text{ms}^{-1}

  • D

    8 ms−18 \, \text{ms}^{-1}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The wave equation is y(x,t)=0.05sin⁡(8x−4t) my(x, t) = 0.05 \sin(8x - 4t) \, \text{m}.

Find: The velocity of the wave.

For a travelling wave of the form

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

we identify k=8 m−1k = 8 \, \text{m}^{-1} and ω=4 rad/s\omega = 4 \, \text{rad/s}.

The wave speed is

v=ωkv = \frac{\omega}{k}

Substituting the given values,

v=48=0.5 ms−1v = \frac{4}{8} = 0.5 \, \text{ms}^{-1}

Therefore, the correct option is C and the velocity of the wave is 0.5 ms−10.5 \, \text{ms}^{-1}.

Common mistakes

  • Using the coefficient of xx or tt directly as the wave speed. This is wrong because those coefficients represent kk and ω\omega, not vv. Use v=ωkv = \frac{\omega}{k} instead.

  • Confusing amplitude 0.05 m0.05 \, \text{m} with wave speed. The amplitude only gives the maximum displacement and has no role in calculating velocity here. Identify kk and ω\omega from the wave equation.

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