PHYS 101 LO4

1. Question:
Two sound waves of frequency 100 Hz travel in the x-direction. One has a displacement amplitude
of 20nm, and the other has a displacement amplitude of 40nm. The phase difference between them
is π/3. What is the resultant amplitude?
Solution:
s_1(x,t)=20cos(kx-ωt)
s_2(x,t)=40cos(kx-ωt+/-π/3)
s_1(x,t)+s_2(x,t)=20cos(kx-ωt)+40cos(kx-ωt+/-π/3)
s_1(x,t)+s_2(x,t)=20cos(kx-ωt)+40cos(kx-ωt)cos(π/3)-/+40sin(kx-ωt)sin(π/3)
s_1(x,t)+s_2(x,t)=40cos(kx-ωt)-/+20√3sin(kx-ωt)————————————————(1)
Assume that s_1(x,t)+s_2(x,t)=Acos(kx-ωt-θ), for some constants A and θ.
So s_1(x,t)+s_2(x,t)=Acos(kx-ωt)cos(θ)+Αsin(kx-ωt)sin(θ)————————(2)
According to equation (1) and (2)
Acos(θ)=40———————————————————————————————(3)
Asin(θ)=-/+20√3—————————————————————————————(4)
Squaring and adding the th6e two equations (3) and (4), we obtain
[Acos(θ)]^2+[Asin(θ)]^2=40^2+[20√3]^2
Α^2[cos^2(θ)+sin^2(θ)]=1600+1200=2800
Α^2=2800
Α=20√7=52.915nm