A cord has two sections with linear densities of 0.12 kg/m and 0.17 kg/m, see th

Question

A cord has two sections with linear densities of 0.12 kg/m and 0.17 kg/m, see the figure. An incident wave, given by D = (0.050 m) sin(11x - 12.0t), where x is in meters and t in seconds, travels along the lighter cord.

(a) What is the wavelength on the lighter section of the cord?
(b) What is the tension in the cord?
(c) What is the wavelength when the wave travels on the heavier section?

For part a, I know that wavelength=2pi/k where k refers to the form D=Asin(kx-wt), which gives me 2pi/11 which equals 0.5712 meters.

How do I calculate parts b and c?

Thanks!

Explanation / Answer

general equation: Asin(kx-wt) a) wavelength1 = 2pi/k = 0.5712 b) Ft does = u*v1^2 you just have to find v1. v1 = w/k = 12/11 = 1.091 u1 = .12 Ft = (.12)(1.091^2 )= .1008 c) now that you've found Ft you can use the *larger linear density* to find the velocity on the thicker part. u2 = .17 v2 = sqrt(Ft/u2). v2 = sqrt(.1008/.17) = .7701 Unfortunately thats not what the question is asking for, we need to find the wavelength on the thicker part. wavelength = T*v2 T = 2pi/w=.5236 so wavelength2 = (.5236)*(.7701) = .4032

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