A transverse periodic wave is represented by the equation y(x, t) = 2.50 cm cos(2,500 rad/s t − 15.0 m⁻¹ x). What are the wavenumber k and direction of propagation of the wave? A) 15.0 m⁻¹; traveling in the − x direction B) 15.0 m⁻¹; traveling in the + x direction C) 30.0 m⁻¹; traveling in the + y direction D) 30.0 m⁻¹; traveling in the − y direction E) 45.0 m⁻¹; traveling in the + x direction

Question

Quizplus · Accepted Answer

The general form of a transverse periodic wave is y(x,t) = A cos(kx - ωt + φ), where A is the amplitude of the wave, k is the wavenumber or wave vector, ω is the angular frequency, t is time, x is position, and φ is the phase angle.

Comparing the given equation to the general form, we can see that:
- A = 2.50 cm (amplitude)
- ω = 2,500 rad/s (angular frequency)

To find the wavenumber and direction of propagation, we need to compare the given equation to the form y(x,t) = A cos(kx - ωt). This means setting the phase angle φ to zero and rearranging the equation to isolate k:

y(x,t) = 2.50 cm cos(2,500 rad/s t − 15.0 mS-1 x)
2.50 cm = A
2,500 rad/s = ω
kx - ωt = -15.0 mS-1 x
kx + 15.0 mS-1 x = ωt
x(k + 15.0 mS-1) = ωt
k + 15.0 mS-1 = ω/t
k = (ω/t) - 15.0 mS-1
k = [2,500 rad/s] / [t]
We need to use the units that make the answer in mS-1 to match the given units. We know that the wave has a period T = 2π/ω, which is the time for one full cycle of the wave. Hence, we have:
T = 2π/ω = 2π/(2,500 rad/s) = 0.00251 s
The wavenumber is then:
k = [2,500 rad/s] / [0.00251 s] - 15.0 mS-1
k = 994.42 m-1 - 15.0 mS-1
k = 979.42 m-1 = 9.79 x 102 m-1
The direction of propagation can be determined from the sign of k. If k is positive, the wave is propagating in the + x direction, and if k is negative, the wave is propagating in the - x direction. Since k is positive in this case, the wave is traveling in the + x direction.

Therefore, the answer is B: 15.0 mS-1, traveling in the + x direction.

A Transverse Periodic Wave Is Represented by the Equation Y(x