How can i calculate interference of waves?

2 Answers
Mar 9, 2018

This is an extremely vague question. I suggest you start by taking a look at the hyperphysics page, as this is probably the level of detail you might need.

The wiki page is actually pretty well detailed on derivations if you require them.

Mar 9, 2018

See below . This should get you started.

Explanation:

Start with two waves of equal amplitude (A) , equal frequency ww, ( expressed in radians { w=2pifw=2πf} but with a phase difference phiϕ

x_1= A coswtx1=Acoswt
x_2= A cos(wt + phi)x2=Acos(wt+ϕ)

Rememeber this identity:

cos(x)+cos(y)=2cos(( x+y)/2)cos((x-y)/2)cos(x)+cos(y)=2cos(x+y2)cos(xy2)

If you add above two waves: resultant wave x_rxr is:

x_r = x_1 + x_2xr=x1+x2
x_r = A coswt + A cos(wt + phi)xr=Acoswt+Acos(wt+ϕ), which can be simplified to

x_r = 2A cos(phi/2)cos(wt + phi/2)xr=2Acos(ϕ2)cos(wt+ϕ2)

Now you can have constructive interference or destructive interference .

For constructive, in above equation, cos(phi/2)=1cos(ϕ2)=1
which gives values for phi = 0,2pi, 4pi etcϕ=0,2π,4πetc

For distructive, in above equation, cos(phi/2)=0cos(ϕ2)=0
which gives values for phi = pi,3pi, 5pi etcϕ=π,3π,5πetc

You can easily plot these values now.

If the frequencies are different, but no pase difference,you would have :

x_1= A cosw_1tx1=Acosw1t
x_2= A cos(w_2t)x2=Acos(w2t)

The resultant superposed wave would be given by this identity:

x_r = 2A cos((w_1 +w_2)/2)cos((w_1 -w_2)/2)xr=2Acos(w1+w22)cos(w1w22)

The result is a wave which is the product of two waves which are the sum and difference of the original waves, so you get something called beats.

If you want an interactive demo check this excellent site:

https://academo.org/demos/amplitude-modulation/