Wave interference and resolving resultants

[kel]

Me again !

I have the following question and this time I really have no idea!

2 sinusoidal waves, identical except for their phase, travel in the same direction along a stretched string and interfere to produce a resultant wave given by:

x(z,t) = 3Cos (20z-4t+0.82)

Where x is in mm, z in metres and t in seconds

a- What is the wavelength of the 2 original waves
b- What is the phase difference between them
c- What is their amplitude

At the present I'm stuck on 'a' and may/may not be able to do b and c once I know how to work out 'a'.

I can see how to add original waves, but how do I go about resolving the resultant wave? I guess it's similar to vector algebra, but mmmmm... really don't know where to begin on this one.

Any help would be very much appreciated.

 

[Päällikkö]

This should get you started:
$$\sin \alpha + \sin \beta = 2 \cos \left( \frac{\alpha - \beta}{2} \right) \sin \left( \frac{\alpha + \beta}{2} \right)$$

 

[kel]

You'll have to excuse me for being a bit clueless here, but how do I relate that to the resultant wave?

Sorry, it's late and my brain needs caffine!:zzz:
Cheers

 

[Päällikkö]

Adding the two waves gives you the resultant wave.

Use the fact that the waves are identical, except for phase, to write equations for the two waves.

 

[kel]

So far, I've got the following

y(z,t) = 3cos(20z-4t) + 3cos(20z-4t+0.82)

but I don't think this is right

 

[Päällikkö]

The two waves, identical except for phase:
$$y(z,t) = A\cos(kz - \omega t + \phi_1) + A\cos(kz - \omega t + \phi_2)$$The earlier equation for cosines:
$$\cos \alpha + \cos \beta = 2 \cos \left( \frac{\alpha - \beta}{2} \right) \cos \left( \frac{\alpha + \beta}{2} \right)$$

Can you now see where we're going?

 

[kel]

So, I'd get

$$\cos \alpha + \cos \beta = 2 \cos \left( \frac{\0 - \0.82}{2} \right) \cos \left( \frac{\0.82}{2} \right)$$
Which leaves
$$\cos \alpha + \cos \beta = 2 \cos \left(-0.41\right) \cos \left(0.41\right)$$

 

[Päällikkö]

I get:
$$y(z,t) = A\cos(kz - \omega t + \phi_1) + A\cos(kz - \omega t + \phi_2)$$
$$y(z,t) = 2A \cos\left(\frac{\phi_1-\phi_2}{2}\right) \cos\left(\frac{2kz - 2\omega t + \phi_1 + \phi_2}{2}\right)$$$$y(z,t) = A' \cos \left( kz - \omega t + \frac{\phi_1 + \phi_2}{2} \right)$$ , where
$$A' = 2A \cos \left(\frac{\phi_1-\phi_2}{2}\right)$$
Now this looks quite a bit like the equation given in the problem.

 

[kel]

I tried it like this (adding two waves together):

a*sin(kx-wt+PI/2) + a*sin(kx-wt+PHI+PI/2) ---> a*cos(kx-wt) +
a*cos(kx-wt+PHI).

adding these using the trig identities you get:
2a*cos(kx-wt+PHI/2)*cos(-PHI/2).

comparing this with the resultant wave, then:
k = 20
w = 4 and
PHI/2 = 0.82.

and using k = 2PI/Lambda and w=2P

wave 2a = 3 this would suggest that the amplitude is 3/2.

 

[Päällikkö]

Good .
The thing I'll have to disagree about is the amplitude. The latter cosine term, cos(-PHI/2) = cos(PHI/2), contributes into the amplitude (see A' in my previous post).