Equation of a Wave Homework: Intensity Variation w/ Phase Difference

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The discussion focuses on deriving how the intensity of two interfering waves varies with their phase difference using the equation I = a². Participants clarify that the double angle formulas and trigonometric identities, such as sin(a) + sin(b), are essential for this derivation. The combined amplitude of the two waves is expressed as a function of their phase difference, leading to an amplitude of 2a°cos(0.5(φ1 - φ2))sin(0.5(2ωt - 2kx + φ1 + φ2)). The conversation emphasizes the importance of understanding the implications of phase differences on intensity, particularly when φ1 equals φ2. Overall, the thread provides insights into the mathematical approach needed to solve the homework problem effectively.
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Homework Statement



Equation of a wave is a = a°sin(ωt - kx + φ)

where φ is the phase of a wave. if 2 waves with phases φ1 and φ2 interfere, show how the intensity I =a² varies as a function of the phase difference φ1 - φ2. Use one of the trigonometric double angle forumula or otherwise to obtain your result.



Homework Equations


The double angle formulas



The Attempt at a Solution



Well am I supposed to map I =a² onto the equation?

If so then the only double angle formula is cos(2x) = cos²(x) - sin²(x)

But I get a really stupid answer when I square the wave equation..

What do I do?
 
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start by writing the sum of the amplitude of the 2 waves with same fequency but different phase & work from there...

also intuitively, what do you expect will happen?
 
I have 2 waves with a phase difference:

a1 = a°sin(ωt - kx + φ1)
a2 = a°sin(ωt - kx + φ2)

If the waves combine, then an interference occurrs...

a1 + a2 = a°sin(ωt - kx + φ1) + a°sin(ωt - kx + φ2)

K apparently, sin a + sin b = 2cos 0.5(a - b) sin 0.5(a + b)

So in relation:

a1 + a2 = 2a°cos 0.5(φ1-φ2) sin 0.5(2ωt - 2kx + φ1 + φ2)

a3 = 2a°cos 0.5(φ1-φ2) sin 0.5(2ωt - 2kx + φ1 + φ2)

Now I =a²

But I'm unsure of how to square this expression I have, assuming it is even right..
 
Unto said:
Use one of the trigonometric double angle forumula or otherwise to obtain your result.

If so then the only double angle formula is cos(2x) = cos²(x) - sin²(x)

Hi Unto! :smile:

I think by a "double angle formula" they mean like sin(A + B) or (sinA + sinB) etc …

these are trigonometric identities which you must learn. :wink:
 
using a few diffenrt trigonamteric identities you can show the identity you used
sin(a) + sin(b) = 2cos(\frac{a-b}{2})sin(\frac{a+b}{2})
Unto said:
a3 = 2a°cos 0.5(φ1-φ2) sin 0.5(2ωt - 2kx + φ1 + φ2)
QUOTE]
so you get the amplitude as
2cos(\frac{\phi_1-\phi_}{2})sin(\omega t - k x \frac{\phi_1+\phi_}{2})
which looks reasonable to me

before you even consider intensity, have a look at what you've got with the amplitude
- the sin term looks similar to the input waves just with a different phase factor
- now the cos term, what happens when phi_1 = phi_2? and for what z, does cos(z) = 0?
- then think about the periodicity of the functions
 
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