A harmonic wave traveling through an initial elastic medium is represented by the wave equ… — JAMB UTME

A harmonic wave traveling through an initial elastic medium is represented by the wave equation $y = 0.04 \cos\left(50\pi t - \frac{\pi}{4} x\right)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. When this wave passes into a second medium, its propagation speed doubles. What is the wavelength of the wave in the second medium?

A
$4.0\text{ m}$
B
$8.0\text{ m}$
$16.0\text{ m}$ Answer
D
$32.0\text{ m}$

Answer

The wavelength of the wave in the second medium is

16.0\text{ m}

The wave's angular frequency

\omega = 50\pi\text{ rad/s}

corresponds to a source frequency of

f = 25\text{ Hz}

, and its wave number

k = \frac{\pi}{4}\text{ rad/m}

corresponds to an initial wavelength

\lambda_1 = 8.0\text{ m}

. The wave speed in the first medium is

v_1 = f \lambda_1 = 200\text{ m/s}

. In the second medium, the wave speed doubles to

v_2 = 400\text{ m/s}

. Because frequency is determined by the source and remains invariant during refraction across media boundaries (

f_2 = f_1 = 25\text{ Hz}

), the new wavelength becomes

\lambda_2 = \frac{v_2}{f} = \frac{400\text{ m/s}}{25\text{ Hz}} = 16.0\text{ m}

Step-by-Step Solution

Extract angular frequency

\omega

and wave number

k

from the general wave equation

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

\omega = 50\pi\text{ rad/s}

and

k = \frac{\pi}{4}\text{ rad/m}

Matching coefficients in the standard wave equation provides the temporal and spatial frequencies of the wave.

Calculate the frequency

f

and wavelength

\lambda_1

in the first medium.

f = \frac{\omega}{2\pi} = \frac{50\pi}{2\pi} = 25\text{ Hz}

and

\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{\pi/4} = 8.0\text{ m}

Fundamental wave relationships connect angular frequency to frequency and wave number to wavelength.

Determine the wave speed

v_1

in the first medium and

v_2

in the second medium.

v_1 = f \lambda_1 = 25 \times 8.0 = 200\text{ m/s}

. Therefore,

v_2 = 2 \times v_1 = 400\text{ m/s}

The problem states that wave propagation speed doubles upon entering the second medium.

Calculate the wavelength

\lambda_2

in the second medium using constant frequency

f = 25\text{ Hz}

\lambda_2 = \frac{v_2}{f} = \frac{400}{25} = 16.0\text{ m}

When a wave crosses a boundary between two media, its frequency is determined solely by the source and remains constant.

Key Concept

Frequency invariance across media boundaries and the wave speed equation

v = f \lambda

Estimated Time:1m 30s

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