A progressive wave traveling through a primary medium is described by the displacement equ… — JAMB UTME

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

$0.375\text{ m}$ Cevap
B
$0.500\text{ m}$
C
$0.188\text{ m}$
D
$2.667\text{ m}$

Cevap

The wavelength of the wave in the secondary medium is

0.375\text{ m}

The option specifying

0.375\text{ m}

is correct because wave frequency

f

remains constant when moving between different media. Extracting

f = \frac{80\pi}{2\pi} = 40\text{ Hz}

from the initial equation allows direct calculation of the new wavelength using

\lambda_2 = \frac{v_2}{f} = \frac{15}{40} = 0.375\text{ m}

Adım Adım Çözüm

Extract angular frequency and wavenumber from the wave equation

From

y = 0.05 \sin(80\pi t - 4\pi x)

, we identify

\omega = 80\pi\text{ rad s}^{-1}

and

k = 4\pi\text{ rad m}^{-1}

The standard wave equation form is

y = A \sin(\omega t - k x)

Determine the constant frequency of the wave

f = \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.

Calculate the wavelength in the secondary medium using the new speed

\lambda_2 = \frac{v_2}{f} = \frac{15\text{ m s}^{-1}}{40\text{ Hz}} = 0.375\text{ m}

The wave speed formula

v = f \lambda

rearranged gives

\lambda = \frac{v}{f}

Anahtar Kavram

Wave speed changes across media boundaries while frequency remains invariant

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