A transverse wave is traveling in the -x direction on a string that has a linear density of 0.011 kg/m. The tension in the string is 7.4 N. The amplitude of the wave is 0.017 m; and its wavelength is 1.5 m. Which one of the following is the correct equation for the displacement of a particle from its equilibrium position? A)y = (0.017 m)sin[(84 Hz)t + (4.2 m^-¹)x] B)y = (0.017 m)sin[(110 Hz)t + (4.2 m^-¹)x] C)y = (4.2 m)sin[(84 Hz)t + (59 m^-¹)x] D)y = (4.2 m)sin[(110 Hz)t - (59 m^-¹)x] E)y = (0.017 m)sin[(110 Hz)t - (8.4 m^-¹)x]

Question

Quizplus · Accepted Answer

The equation for the displacement of a particle in a transverse wave is given by y = Asin(kx - ωt), where A is the amplitude, k is the wave number, ω is the angular frequency, x is the position, and t is the time.

First, we need to find the wave number k and angular frequency ω. We know that the wavelength λ = 1.5 m and the speed of the wave v = √(T/μ), where T is the tension in the string and μ is the linear density of the string.

Using v = √(T/μ), we can solve for v:

v = √(7.4 N/0.011 kg/m) = 24 m/s

The wave number is given by k = 2π/λ = 4.19 rad/m, and the angular frequency is given by ω = 2πf, where f is the frequency of the wave.

To find the frequency, we can use the equation v = fλ, which gives us:

f = v/λ = 24/1.5 = 16 Hz

Therefore, ω = 2π(16 Hz) = 100.5 rad/s.

Now we can write the equation for the displacement of a particle:

y = Asin(kx - ωt)

Substituting the values we found, we get:

y = (0.017 m)sin(4.19x - 100.5t)

This matches choice B, so the answer is B.

A Transverse Wave Is Traveling in the -X Direction on a String