How Do You Find the Combined Elongation and Phase Shift of Superimposed Waves?

[Gauss M.D.]

Homework Statement

Find the combined elongation of the waves 7sin(wt) and 2sin(wt + pi/4). Express it both in real and complex form.

Homework Equations

A = sqrt(A₁² + A₂² + A₁A₂cos(Δθ)

The Attempt at a Solution

I was given the formula above, which I don't understand, but it does spit out the correct amplitude (8.53). But how do I find the phase shift of the new wave? I tried:

g(t) = 7sin(wt) + 2si(wt + pi/4)
g(0) = 2sin(pi/4) = √2

f(t) = 8.53sin(wt + θ)
f(0) = 8.53sin(θ)

But g(0) = f(0) since f(t) = g(t) so 8.53sinθ = √2

arcsin(√2/8.52) = θ = 0.167

Which is incorrect. How do I find the phase of the new wave?

 

[Gauss M.D.]

I'm a moron ty

 

[barryj]

What is meant by elongation? The formula you give looks like the law of cosines. It seems that you are basically vectorically adding the two waves.

BTW, how did you get the answer of 8.53?

 

[ehild]

Homework Statement

Find the combined elongation of the waves 7sin(wt) and 2sin(wt + pi/4). Express it both in real and complex form.

Homework Equations

A = sqrt(A₁² + A₂² + A₁A₂cos(Δθ)

The Attempt at a Solution

I was given the formula above, which I don't understand, but it does spit out the correct amplitude (8.53).

Adding two SHM-s with the same angular frequency, f1= asin(wt) and f2=bsin(wt+ψ), it results is a single SHM, with amplitude C and phase constant θ:

g(t)=asin(wt)+bsin(wt+ψ)=Csin(wt+θ)

Applying the addition rule, expand sin(wt+ψ) and sin(wt+θ).

asin(wt)+b[sin(wt)cosψ+cos(wt)sinψ)=C[sin(wt)cosθ+cos(wt)sinθ]

Collect the terms containing sin(wt) and cos(wt):

sin(wt)[a+bcosψ]+bcos(wt)sinψ=Csin(wt)cosθ+Ccos(wt)sinθ

The equation is an identity, it must hold for every value of wt.

If wt = 0 cos(wt) = 1: bsinψ=Csinθ *

When wt=pi/2 sin(wt)=1, cos(wt)=0: a+bcosψ=Ccosθ. **

Square both equations and add them: b²sin²ψ+(a²+2abcosψ+b²cos²ψ)=C²[sin²wt)+cos²(wt)]

Use that sin²ψ+cos²ψ=1

a²+b²+2abcosψ = C²

Dividing the equations * and **

tanθ=sinψ/(a+bcosψ).

You can decide about the quadrants from the sign of sinθ =bsinψ/C, cosθ=(a+bcosψ)/C.

Your solution was correct.

ehild

 

[barryj]

I didn't mean that Gauss MD's answer was incorrect but if he plugged his numbers into the equation he gave you would get a different answer. He is missing a 2 I believe in the formula. Also, some insight might be helpful. If you have any number of waves, sinusoidal, they can be added vectorically. The answer can easily be achieved by not using the formula but instead doing basic vector addition. Also, the angle is the angle of the resulting vector. If he had 100 waves, added together, at the same frequency, the result could be obtained using simple vector addition that is normally learned in Geometry or Alg 2. In electrical engineering, we often refer to the sinusoids as a phasors represented in this case by 2/_ pi/4 and 7/_ 0. From a mathematical point of view using trig identities, ehild's derivation is correct.

Also some insight, this is still a variation of the law of cosines, from geometry. If you draw the vectors, pi/4 becomes 3pi/4 as the included angle between the vectors and the law of cosines will give the resultant of the sum.

 

[ehild]

I didn't mean that Gauss MD's answer was incorrect but if he plugged his numbers into the equation he gave you would get a different answer. He is missing a 2 I believe in the formula. A

Yes his equation was wrong.

ehild