Question

Difficulty: MediumWave Properties and Mathematical Wave Equation

A progressive wave traveling through a primary medium is described by the displacement equation y=0.05sin(80πt4πx)y = 0.05 \sin(80\pi t - 4\pi x), where xx and yy are in meters and tt is in seconds. If the wave enters a secondary medium where its speed decreases to 15 m s115\text{ m s}^{-1}, what is the wavelength of the wave in the secondary medium?

  1. 0.375 m0.375\text{ m}Answer
  2. B
    0.500 m0.500\text{ m}
  3. C
    0.188 m0.188\text{ m}
  4. D
    2.667 m2.667\text{ m}

Answer

The wavelength of the wave in the secondary medium is 0.375 m0.375\text{ m}.
The option specifying 0.375 m0.375\text{ m} is correct because wave frequency ff remains constant when moving between different media. Extracting f=80π2π=40 Hzf = \frac{80\pi}{2\pi} = 40\text{ Hz} from the initial equation allows direct calculation of the new wavelength using λ2=v2f=1540=0.375 m\lambda_2 = \frac{v_2}{f} = \frac{15}{40} = 0.375\text{ m}.

Step-by-Step Solution

1
Extract angular frequency and wavenumber from the wave equation
From y=0.05sin(80πt4πx)y = 0.05 \sin(80\pi t - 4\pi x), we identify ω=80π rad s1\omega = 80\pi\text{ rad s}^{-1} and k=4π rad m1k = 4\pi\text{ rad m}^{-1}.
The standard wave equation form is y=Asin(ωtkx)y = A \sin(\omega t - k x).
2
Determine the constant frequency of the wave
f=ω2π=80π2π=40 Hzf = \frac{\omega}{2\pi} = \frac{80\pi}{2\pi} = 40\text{ Hz}.
Frequency depends solely on the wave source and remains unchanged when crossing media boundaries.
3
Calculate the wavelength in the secondary medium using the new speed
λ2=v2f=15 m s140 Hz=0.375 m\lambda_2 = \frac{v_2}{f} = \frac{15\text{ m s}^{-1}}{40\text{ Hz}} = 0.375\text{ m}.
The wave speed formula v=fλv = f \lambda rearranged gives λ=vf\lambda = \frac{v}{f}.

Key Concept

Wave speed changes across media boundaries while frequency remains invariant
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