Standing Waves

[nycmathguy]

Section 5.4

 

[MathLover1]

y[1]=A*cos(2pi(t/T-x/λ),

y[2]=A*cos(2pi(t/T +x/λ),

y[1]+y[2]=A*cos(2pi(t/T-x/λ))+A*cos(2pi(t/T +x/λ)

y[1]+y[2]=A(cos((2πt)/T - (2πx)/λ)+cos((2πt)/T + (2 π x)/λ))

y[1]+y[2]=A(2cos((2πt)/T) cos((2πx)/λ))

y[1]+y[2]=2A(cos((2πt)/T) cos((2πx)/λ))

 

[nycmathguy]

y[1]=A*cos(2pi(t/T-x/λ),

y[2]=A*cos(2pi(t/T +x/λ),

y[1]+y[2]=A*cos(2pi(t/T-x/λ))+A*cos(2pi(t/T +x/λ)

y[1]+y[2]=A(cos((2πt)/T - (2πx)/λ)+cos((2πt)/T + (2 π x)/λ))

y[1]+y[2]=A(2cos((2πt)/T) cos((2πx)/λ))

y[1]+y[2]=2A(cos((2πt)/T) cos((2πx)/λ))

1. Did you skip any steps?

2. How did you get
y[1]+y[2]=A(2cos((2πt)/T) cos((2πx)/λ))?

 

[MathLover1]

y[1]+y[2]=A*cos(2pi(t/T-x/λ))+A*cos(2pi(t/T +x/λ).......factor out A
y[1]+y[2]=A(cos(2pi(t/T-x/λ))+cos(2pi(t/T +x/λ))

using cos(u-v) and cos(u+v) you prove that

then you have

 

[nycmathguy]

y[1]+y[2]=A*cos(2pi(t/T-x/λ))+A*cos(2pi(t/T +x/λ).......factor out A
y[1]+y[2]=A(cos(2pi(t/T-x/λ))+cos(2pi(t/T +x/λ))

using cos(u-v) and cos(u+v) you prove that

then you have

Originally, I thought about using cos(u - v) and
cos(u + v) but was not sure what to do with A.
Factoring comes in handy after prealgebra.