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The velocity of the plane wave sin2(ωt - βx) isa) 2ω / βb) ω / 2βc) ω2 / β2d) ω / βSelect one:badc

Question

The velocity of the plane wave sin2(ωt - βx) is

a) 2ω / β
b) ω / 2β
c) ω² / β²
d) ω / β

Select one:
badc

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Solution

To determine the velocity of the plane wave given by the equation sin⁡(2(ωt−βx)) \sin(2(\omega t - \beta x)) , we can analyze the wave function.

1. Break Down the Problem

Identify the parameters in the wave equation and what they represent:

  • The term ω \omega refers to the angular frequency.
  • The term β \beta is the wave number.
  • The form of the wave equation helps us find the wave velocity.

2. Relevant Concepts

The general form of a plane wave can be described as Asin⁡(ωt−kx) A \sin(\omega t - kx) , where:

  • k k is the wave number, which in this case is β \beta .
  • The velocity v v of the wave is given by the formula: v=ωk v = \frac{\omega}{k}

3. Analysis and Detail

Substituting k k with β \beta : v=ωβ v = \frac{\omega}{\beta}

4. Verify and Summarize

From our analysis, the wave velocity derived from the equation is consistent with the standard wave velocity formula.

Final Answer

The velocity of the plane wave sin⁡(2(ωt−βx)) \sin(2(\omega t - \beta x)) is ωβ \frac{\omega}{\beta} . Thus, the correct answer is d) ωβ\frac{\omega}{\beta}.

This problem has been solved

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