Asked by Kid

A cord has two sections with linear densities of 0.10 kg/m and 0.20 kg/m respectively. An incident wave, y(x,t) = 0.050sin(6.0x - 12.0t), where x is in meters and t in seconds, travels from the lighter cord to the heavier one. What is the wavelength of the wave in each section of the cord?

So I interpret the wave going through like this:

|--------1--------|--------2--------|

In cord 1, the linear density is u_1 = 0.10kg/m. In cord 2, the density u_2 = 0.20kg/m.

The wave goes through cord 1 first, and then through cord 2.

Using the template y(x,t) = Asin(kx - wt) and from the given function y(x,t) = 0.050sin(6.0x - 12.0t):

k = 6.0 m^-1
w = 12.0 rad/s
k = 2pi/lambda -> lambda = 2pi/k = (2)(pi)/6.0m^-1 = 1.047197551m = 1.05m

And I guess that lambda_1 = 1.05m then?
But how do I get lambda_2 (the wavelength of the wave travelling through cord 2)?

Answers

Answered by Damon
The frequency is the same on both sides of the transition. There is no box to store waves in so as many have to pass a point in an hour on the left as on the right. Same f and same w and same 1/f = period T

The tension is the same both sides

w = 2 pi f = the same both sides
the s

in the light line
12/f = 2 pi
so f = 6/pi and w = 2 pi f = 12rad/s agreed
for wavelength
6.0 lambda = 2 pi
lambda = pi/3 = 1.05 sure enough

NOW
Double the mass per unit length
speed of transverse wave = sqrt(tension/mass per unit length)
so
the speed is the original speed /sqrt 2
so in the same amount of time (period) it goes original lambda/sqrt 2
= 1.05/sqrt 2
Answered by Kid
How did you get "sqrt 2" from "sqrt(tension/mass per unit length)"?

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