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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

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Solution

To determine the velocity of the plane wave given by the equation sin(2(ωtβx)) \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(ωtkx) 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(ωtβx)) \sin(2(\omega t - \beta x)) is ωβ \frac{\omega}{\beta} . Thus, the correct answer is d) ωβ\frac{\omega}{\beta}.

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