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?

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

Cevap

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}

Adım Adım Çözüm

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.

Anahtar Kavram

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

Soru

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Konular

Bu konudan daha fazla soru