A sinusoidal progressive wave propagating through an elastic medium is described by the eq… — JAMB UTME

A sinusoidal progressive wave propagating through an elastic medium is described by the equation $y = 0.04 \sin(60\pi t - 4\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. As the wave enters a second medium, its wave speed increases by $50\%$ . What is the wavelength of the wave in the second medium?

A
$0.33\text{ m}$
B
$0.50\text{ m}$
$0.75\text{ m}$ Answer
D
$1.50\text{ m}$

Answer

The wavelength of the wave in the second medium is

0.75\text{ m}

Comparing

y = 0.04 \sin(60\pi t - 4\pi x)

to the standard wave equation

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

gives

\omega = 60\pi\text{ rad/s}

and

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

. The source frequency is

f = \frac{\omega}{2\pi} = 30\text{ Hz}

, and the initial wavelength is

\lambda_1 = \frac{2\pi}{k} = 0.50\text{ m}

. The initial wave speed is

v_1 = \frac{\omega}{k} = 15\text{ m/s}

. When the wave enters the second medium, its speed increases by

50\%

v_2 = 22.5\text{ m/s}

. Frequency is a source property and remains constant (

30\text{ Hz}

) across media boundaries, so the new wavelength becomes

\lambda_2 = \frac{v_2}{f} = \frac{22.5}{30} = 0.75\text{ m}

Step-by-Step Solution

Extract angular frequency and wave number from the progressive wave equation.

From

y = 0.04 \sin(60\pi t - 4\pi x)

, we find

\omega = 60\pi\text{ rad/s}

and

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

Standard form of a harmonic progressive wave is

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

Calculate the frequency and initial wavelength in the first medium.

Frequency

f = \frac{\omega}{2\pi} = \frac{60\pi}{2\pi} = 30\text{ Hz}

. Initial wavelength

\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{4\pi} = 0.50\text{ m}

Wave frequency and wavelength are related to angular parameters by

\omega = 2\pi f

and

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

Determine the initial wave speed and the new speed in the second medium.

Initial speed

v_1 = f \lambda_1 = 30 \times 0.50 = 15\text{ m/s}

. New speed

v_2 = 15 \times (1 + 0.50) = 22.5\text{ m/s}

Wave speed increases by

50\%

upon entering the second medium.

Apply boundary condition for wave refraction to determine the new wavelength.

Since frequency is invariant across media boundaries (

f_2 = f_1 = 30\text{ Hz}

\lambda_2 = \frac{v_2}{f_2} = \frac{22.5}{30} = 0.75\text{ m}

Frequency depends solely on the wave source and does not change when passing between media.

Key Concept

Frequency invariance and wavelength alteration across media boundaries in progressive waves

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