# Calculating resultant amplitude, given 2 ampli's and phase diff

1. Jul 22, 2011

### RKOwens4

1. The problem statement, all variables and given/known data

Two waves of the same frequency have amplitudes 1.02 and 2.27. They interfere at a point where their phase difference is 59.5°. What is the resultant amplitude?

2. Relevant equations

Resultant Amp = (Amp1 + Amp2)cos(theta/2)

3. The attempt at a solution

Okay this seemed like a simple plug and chug problem. I ended up getting an answer of 2.86, but this is marked wrong. I then did the "practice another version" thing on webassign and worked another version of the same exact problem (just with different values of the two amps and a different theta), but used the exact same formula and guess what... it was correct! I've spent the last 15 minutes trying to figure out why the formula works for one version of the problem but not the one that matters. Can anyone help?

2. Jul 22, 2011

### ehild

Are you sure that this formula is true?

ehild

3. Jul 22, 2011

### Staff: Mentor

It looks like it holds true when the frequencies of the two signals are identical.

4. Jul 22, 2011

### RKOwens4

Can someone help, please? I don't expect anyone to just solve the problem and give me the answer like they do on cramster, but statements like the two above (with all due respect) don't really help me one single iota. No, I'm not sure if that equation works 100% of the time (obviously, it doesn't). But, what formula should I try instead?

5. Jul 22, 2011

### Staff: Mentor

The formula is correct. If your result, 2.856, or rounded to two sig figs, 2.86 is marked incorrect, then either:

1. There are some units that should have been specified
2. There is a typo in the question (no way to fix that!)
3. The marking program is incorrectly programmed for this question (call your instructor)

6. Jul 22, 2011

### ehild

Try the following formula for the resultant amplitude :

A=sqrt[A11+A22-2A1A2cos(θ)].

ehild

7. Jul 22, 2011

### Staff: Mentor

#### Attached Files:

• ###### Fig1.gif
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8. Jul 22, 2011

### ehild

The interfering waves are A1sin(wt-kx) and A2sin(wt-kx+θ). Their resultant is a single wave with amplitude A. Denoting wt-kx by α, we have the sum

A1sin(α)+A2sin(α+θ)=Asin(α+φ)*.

sin(α+θ)=sin(α)cos(θ)+cos(α)sin(θ), and sin(α+φ)=sin(α)cos(φ)+cos(α)sin(φ).

Replacing into eq. * :

sin(α)[A1+A2cos(θ)]+cos(α)A2sin(θ)=A[sin(α)cos(φ)+cos(α)sin(φ)]

The equation is valid for any angles, that is for any values of sin(α) and cos(α) between -1 and 1. Therefore

A1+A2cos(θ)=Acos(φ) and A2sin(θ)=Asin(φ)

Square both equation and add up:

A2=A12+A22+2A1A2cos(θ).

ehild

9. Jul 22, 2011

### ehild

Let be the phase difference θ=pi. Then the resultant amplitude is always 0, according to the formula in the original post which is certainly not true when the amplitudes are different.

ehild

10. Jul 22, 2011

### ehild

Is the resultant amplitude 2.86, according to the plot? It is close but different.

ehild

11. Jul 22, 2011

### Staff: Mentor

I've been looking more closely at the OP's formula. It would appear to be a good approximation for small phase angles, say less than 50 degrees or so. I'm currently investigating a derivation that will provide an error term.

12. Jul 22, 2011

### Staff: Mentor

The plot maxima are at amplitude 2.92 ! So yes, close to 2.86, but not spot on. Your formula delivers the goods (as it should, based upon the derivation!).

13. Jul 22, 2011

### ehild

I wonder why not the exact formula is used? It is derived in every books on General Physics.

ehild

14. Jul 22, 2011

### Staff: Mentor

I suppose it's one of those shortcuts that engineers use
Makes the math simpler.

I've derived an expression for the signal sum that looks as follows:

$$(A + B) cos\left(\frac{\phi}{2}\right) cos(\omega t) + (A - B)sin\left(\frac{\phi}{2}\right) sin(\omega t)$$

The amplitude of the left term is clearly the OP's formula. The right term represents a deviation from that formula. Apparently the OP's formula works best for similar amplitudes and small angles.

15. Jul 22, 2011

### ehild

I see now where the OP's formula came from. Thanks gneil! But it is really a rough approximation.

ehild

16. Jul 22, 2011

### Staff: Mentor

I can't argue with you there!

17. Jul 22, 2011

### ehild

What if students try to use the resultant for A1=3 A2=1 and phase difference of 180°? They see only a formula to plug in data. Applying the correct formula is not much more work than using the wrong one.Why do they teach the wrong one then?

ehild

18. Jul 23, 2011

### Staff: Mentor

To be fair, we don't know the background of where that formula comes from. For all we know it was a perfectly valid formula in a particular example (equal amplitude waves, perhaps), that was lifted out of context. We would have to as the OP where he found it.