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Zorluk: ZorWave Properties and Mathematical Wave Equation

A progressive wave traveling through a primary medium is represented by the displacement equation y=0.04sin(100πt2.5πx)y = 0.04 \sin\left(100\pi t - 2.5\pi x\right), where xx and yy are in meters and tt is in seconds. If the wave propagates into a secondary medium where its speed increases by 20%20\%, what is the wavelength of the wave in the secondary medium?

  1. 0.96 m0.96\text{ m}Cevap
  2. B
    0.67 m0.67\text{ m}
  3. C
    1.50 m1.50\text{ m}
  4. D
    0.16 m0.16\text{ m}

Cevap

The wavelength of the wave in the secondary medium is 0.96 m0.96\text{ m}.
Comparing y=0.04sin(100πt2.5πx)y = 0.04 \sin(100\pi t - 2.5\pi x) to the standard form y=Asin(ωtkx)y = A \sin(\omega t - k x), the wave number is k=2.5π rad/mk = 2.5\pi\text{ rad/m}. The initial wavelength is λ1=2π2.5π=0.8 m\lambda_1 = \frac{2\pi}{2.5\pi} = 0.8\text{ m}. When a wave moves to a new medium, its frequency stays constant, making wavelength directly proportional to wave speed (vλv \propto \lambda). An increase of 20%20\% in wave speed means the new wavelength is λ2=0.8×1.20=0.96 m\lambda_2 = 0.8 \times 1.20 = 0.96\text{ m}.

Adım Adım Çözüm

1
Extract angular frequency ω\omega and wave number kk from the wave equation.
From y=Asin(ωtkx)y = A \sin(\omega t - k x), we identify ω=100π rad/s\omega = 100\pi\text{ rad/s} and k=2.5π rad/mk = 2.5\pi\text{ rad/m}.
Standard wave equation parameters directly define the wave's spatial and temporal frequencies.
2
Calculate the wavelength λ1\lambda_1 in the initial medium.
\(\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{2.5\pi} = 0.8\text{ m}\).
Wavelength is inversely related to the wave number kk by λ=2πk\lambda = \frac{2\pi}{k}.
3
Apply the boundary conditions of wave refraction across media.
Frequency ff remains constant across boundary; speed vv and wavelength λ\lambda scale proportionally.
The frequency of a wave is determined solely by the source and does not change upon entering a new medium.
4
Determine the new wavelength λ2\lambda_2 in the secondary medium.
\(\lambda_2 = \lambda_1 \times (1 + 0.20) = 0.8\text{ m} \times 1.20 = 0.96\text{ m}\).
Since v=fλv = f \lambda and ff is constant, v2v1=λ2λ1=1.20\frac{v_2}{v_1} = \frac{\lambda_2}{\lambda_1} = 1.20.

Anahtar Kavram

Invariance of Wave Frequency across Media and Wave Equation Parameter Extraction
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