Two wires with different linear mass densities are soldered together, end to end
ID: 1444288 • Letter: T
Question
Two wires with different linear mass densities are soldered together, end to end, and are stretched under a tension of F=500N. The wave speed in the second wire is 3 times that in the first wire.
a) What is the ratio between the linear mass density of the first and second wire?
b) What is the ratio between the wavelengths of the waves propagating in each wire?
c) Assuming that at the connection between the two wires one quarter of the power of the incident wave is reflected, and the rest is transmitted, find the amplitude of the reflected wave and the amplitude of the transmitted wave. The amplitude of the incident wave is A=1 cm.
Explanation / Answer
a) wave speed, v = sqrt(T / linear density )
v^2 = T / linear density
v^2 * linear density = T = constant (for this case as T = 500N for both)
v2 = 3v1
v2^2 * lambda2 = v1^2 lambda1
(3v1)^2 lambda2 = v1^2 lambda1
lambda1 / lambd2 = 3^2 = 9
b)
wavelength = speed / frequency
frequency is same.
hence wavelength1 /wavelength2 = v1/v2 = 1/3
c) Power is proportional to amplitude squared.
reflected wave amplitude = 1 x sqrt(1/4) = 0.5 cm
transmitted wave apmlitude = 1 x sqrt(1 - 1/4) = 0.866 cm
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