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
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)?
2 answers
How did you get "sqrt 2" from "sqrt(tension/mass per unit length)"?
0 answers