A wave traveling through a first medium is described by the displacement equation y = 0.02… — JAMB UTME

A wave traveling through a first medium is described by the displacement equation $y = 0.02 \sin \left(100\pi t - \frac{4\pi}{3} x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. As the wave enters a second medium, its speed becomes $120\text{ m/s}$ . What is the wavelength of the wave in the second medium?

$2.4\text{ m}$ Answer
B
$1.5\text{ m}$
C
$0.42\text{ m}$
D
$3.0\text{ m}$

Answer

The wavelength of the wave in the second medium is

2.4\text{ m}

Comparing the wave equation

y = 0.02 \sin\left(100\pi t - \frac{4\pi}{3} x\right)

with

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

gives

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

. The frequency is

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

. Because frequency does not change when crossing boundaries, the wavelength in the second medium where speed is

120\text{ m/s}

\lambda = \frac{v}{f} = \frac{120}{50} = 2.4\text{ m}

Step-by-Step Solution

Extract the angular frequency

\omega

from the given wave equation.

The angular frequency

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

The standard wave equation is

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

Calculate the frequency of the wave.

f = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50\text{ Hz}

Frequency is determined by the source and remains constant regardless of the medium.

Determine the wavelength in the second medium using the new wave speed.

\lambda_2 = \frac{v_2}{f} = \frac{120\text{ m/s}}{50\text{ Hz}} = 2.4\text{ m}

Applying the wave equation

v = f\lambda

with the updated speed in the second medium.

Key Concept

Invariance of wave frequency across media boundaries and extraction of wave parameters from the mathematical wave equation.

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