A progressive wave propagating through an initial medium is represented by the displacemen… — JAMB UTME

A progressive wave propagating through an initial medium is represented by the displacement equation $y = 0.05 \sin\left(100\pi t - \frac{2\pi}{5} x\right)$ , where $x$ and $y$ are in metres and $t$ is in seconds. Upon entering a second medium, the wave speed decreases by $20\%$ . What is the wavelength of the wave in the second medium?

$4.0\text{ m}$ Answer
B
$5.0\text{ m}$
C
$6.25\text{ m}$
D
$2.0\text{ m}$

Answer

4.0\text{ m}

Comparing

y = 0.05 \sin\left(100\pi t - \frac{2\pi}{5} x\right)

with the standard wave equation

y = A \sin(\omega t - kx)

yields a wave number

k = \frac{2\pi}{5}\text{ rad/m}

, giving an initial wavelength

\lambda_1 = \frac{2\pi}{k} = 5.0\text{ m}

. When a wave refracts into another medium, its frequency stays constant, making wave speed and wavelength directly proportional. A

20\%

reduction in speed decreases the wavelength by

20\%

, giving

\lambda_2 = 5.0 \times 0.80 = 4.0\text{ m}

Step-by-Step Solution

Extract the angular frequency

\omega

and wave number

k

from the wave equation

\omega = 100\pi\text{ rad/s}

and

k = \frac{2\pi}{5}\text{ rad/m}

The standard progressive wave equation is

y = A \sin(\omega t - kx)

Calculate the wavelength

\lambda_1

in the first medium

$\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{\frac{2\pi}{5}} = 5.0\text{ m}$

The wave number is related to wavelength by

k = \frac{2\pi}{\lambda}

Determine the new wavelength

\lambda_2

in the second medium using the refraction property

$\lambda_2 = \lambda_1 \times (1 - 0.20) = 5.0 \times 0.80 = 4.0\text{ m}$

When a wave crosses a boundary between two media, its frequency

f

remains constant. Therefore, wave speed

v = f\lambda

is directly proportional to wavelength

\lambda

Key Concept

Constancy of wave frequency during refraction and wave equation parameters extraction

Estimated Time:2m 0s

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