Transverse traveling wave on a cord

[gmmstr827]

Homework Statement

A transverse traveling wave on a cord is represented by D = 0.22sin(5.6x+34t) where D and x are in meters and t is in seconds. For this wave determine (a) the wavelength, (b) frequency, (c) velocity (magnitude and direction), (d) amplitude, and (e) maximum and minimum speeds of particles on the cord.

d)What is the ratio of density for the different sections of a rope (thin on the left with downward reflected pulse, thick on the right with upward (like a wave) transmitted pulse) to create a 1/2*[wavelength] on the right?

Homework Equations

D(x,t) = Asin(kx-εt)
D = 0.22sin(5.6x+34t) where D and x are in meters and t is in seconds
v = Γ/T
T = 1/f
Γ = 2L/n where n = 1 represents fundamental; n = 2 is the second harmonic; etc.

The Attempt at a Solution

Looking at the base formula for a standing wave, I have concluded the following:

a) The wavelength ε = 34 m
b) The frequency f = 5.6 Hz
c) v = Γ/T
T = 1/f = 1/5.6Hz = 0.179 s
Γ = 2L/n >>> Would we assume n = 1? Would we also assume that Length = A = 0.22?
If so:
Γ = 2(0.22) = 0.44 m
v = 0.44/0.179 = 2.46 m/s to the right (since Amplitude is positive and it is only a sine function) (how would I find magnitude?)
d) A = 0.22 m
e) Would this be ±v = ±2.46 m/s?
d) How do I go about this?

Thank you for any help!

 

[anathema]

First, let's clarify the given information. The equation D = 0.22sin(5.6x+34t) represents a traveling wave, not a standing wave. The equation for a standing wave would be D(x,t) = Asin(kx-εt), where k is the wave number and ε is the angular frequency.

a) The wavelength (λ) can be found by using the formula λ = 2π/k. In this case, k = 5.6, so λ = 2π/5.6 = 1.12 m.

b) The frequency (f) is given by the coefficient of t in the equation, which is 34. So, f = 34 Hz.

c) The wave velocity (v) can be found by using the formula v = λf. In this case, v = 1.12 m * 34 Hz = 38.08 m/s to the right.

d) The amplitude (A) is given in the equation as 0.22 m.

e) The maximum and minimum speeds of the particles on the cord can be found by differentiating the equation with respect to time and setting it equal to zero. This gives us the maximum speed as v_max = |Aω| = |0.22*34| = 7.48 m/s and the minimum speed as v_min = -|Aω| = -7.48 m/s.

To answer the second part of the question, we need to use the formula for the velocity of a wave in a string, which is v = √(T/μ), where T is the tension in the string and μ is the linear density. We can rearrange this formula to solve for μ: μ = T/v^2. Since we want the ratio of densities, we can divide the two equations to get μ_left/μ_right = (T_left/T_right)/(v_left^2/v_right^2). From the given information, we know that the wavelength on the right side is half of the wavelength on the left, so v_left = 2v_right. Plugging this into the equation, we get μ_left/μ_right = (T_left/T_right)/(4v_right^2/v_right^2) = (T_left/T_right)/4. Since we want the left side to have half the wavelength, we can set T_left = T_right/2.