Find how many wavelengths of a wave are in a section of wire

[Epa06]

1. A string with linear density 2.0 g/m is strecthed along the positive x-axis with tension 20N. One end of the string, at x=0m, is tied to a hook that oscillates up and down with a frequency of 100 Hz and a max displacement of 1.0mm. AT t=0s, the hook is at its lowest point.
a) What are the wave speed on the string and the wavelength?
b) what are the amplitude and phase constant of the wave?
c) write the equation for the displacement D(x,t) of the traveling wave.
d) what is the strings displacement at x=0.50 m and t=15ms?


2. v=lambda*f, w=2pi*f



3. I'm just confused about the whole problem. I don't understand how the tension plays into the whole thing.

 

[jbsteele]

a) The wave speed on the string is given by v = √(T/μ), where T is the tension and μ is the linear density of the string. Thus, for this problem, the wave speed is v =√(20N/2g/m) = 10 m/s. The wavelength is then given by λ = v/f, so the wavelength is λ = 10 m/s/100Hz = 0.1 m. b) The amplitude of the wave is given by the maximum displacement of the hook, which is 1.0 mm. The phase constant is not given in the problem, so it cannot be determined. c) The equation for the displacement D(x,t) of the traveling wave is given by D(x,t) = A*sin(ωt - kx + φ), where A is the amplitude, ω is the angular frequency, k is the wave number, and φ is the phase constant. Thus, the equation for this problem is D(x,t) = 1.0mm*sin(2π*100Hz*t - 2π/0.1m*x). d) The displacement of the string at x=0.50 m and t=15ms is given by D(0.50m, 15ms) = 1.0mm*sin(2π*100Hz*15ms - 2π/0.1m*0.50m) = 0.94mm.