Solve Superposition Wave: Prove, Phase Angle & Amplitude

[Const@ntine]

Homework Statement

y₁(x,t) = 5.00sin(2.00x - 10.0t)

y₂(x,t) = 10.0cos(2.00x - 10.0t)

a) Prove that the wave that is the result of the superposition is a function of sin.

b) What's the phase angle and amplitude of said wave?

Homework Equations

y = y₁ + y₂

The Attempt at a Solution

Initially I figured I'd work y₂ into a sin function like this: y₂(x,t) = 10.0sin(2.00x - 10.0t + π/2). Then take the y = y₁ + y₂ formula:

y = 5.00(sin(2.00x - 10.00t) + 2sin(2.00x - 10.0t + π/2))

Then I'd set 2.00x - 10.0t = a & π/2 = b, and rework the latter with: sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b)

But if I put that in the above equation, I just end up with y = 5.00sin(2.00x - 10.0t) + 10.0cos(2.00x - 10.0t)

I checked around to find any formulas or theory about waves without the same amplitudes and whatnot, but my book has only the one case (same direction, same A, sin).

Any help is appreciated!

 

[Orodruin]

Since both the sine and cosine have the same phase function, you can make the ansatz $y = A\sin(2x-10t+\phi)$. You can then use the trigonometric identities you mentioned to find the constants $A$ and $\phi$.

 

[Const@ntine]

Since both the sine and cosine have the same phase function, you can make the ansatz $y = A\sin(2x-10t+\phi)$. You can then use the trigonometric identities you mentioned to find the constants $A$ and $\phi$.

Sorry for the extremely late reply (I marched on to thermodynamics and forgot about it), but when you say "trigonomic identities" you mean sin(a+b) = ... ? Apart from that, (a) says that I have to prove the wave is a sin function, so don't know if I could make an ansatz.

 

[Orodruin]

Sorry for the extremely late reply (I marched on to thermodynamics and forgot about it), but when you say "trigonomic identities" you mean sin(a+b) = ... ? Apart from that, (a) says that I have to prove the wave is a sin function, so don't know if I could make an ansatz.

Of course you can make an ansatz if you later show that it is true and identify the constants.

 

[Const@ntine]

Of course you can make an ansatz if you later show that it is true and identify the constants.

I guess. It's just that it broke down the exercise into two questions, and I figured I had to do it this one way, and no other. I'm still kinda stuck on it though. Let's say I make the above ansatz, set (2x - 10t) = a & φ = b, and then follow with:

sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b) = sin(2x - 10t)*cos(φ) + sin(φ)*cos(2x-10t)

What then?

 

[Orodruin]

What then?

Then you figure out what values $\phi$ and $A$ have to take for the expression to be identical to your superposition.

 

[Const@ntine]

Then you figure out what values $\phi$ and $A$ have to take for the expression to be identical to your superposition.

Don't I have to connect the above equation with the info I have? I don't see how I can do that though. I have this equation as well (y = y₁ + y₂):

y = 5.00sin(2.00x - 10.00t) + 10.0cos(2.00x - 10.0t)

But I'm not seeing how I can connect the two.

 

[Orodruin]

But this is exactly what I have been saying. You put them equal to each other and start identifying the terms.

 

[Const@ntine]

But this is exactly what I have been saying. You put them equal to each other and start identifying the terms.

Oh yeah, darn it. It flew right past me.

Asin((2x - 10t) + φ) = 5.00sin(2.00x - 10.00t) + 10.0cos(2.00x - 10.0t)

Asin(a+b) = A*[sin(a)*cos(b) + cos(a)*sin(b) = sin(2x - 10t)*cos(φ) + sin(φ)*cos(2x-10t)]

So:

Asinφ = 10
Acosφ = 5
______________

tanφ = 2 => φ = 1.12 rad = 6.4 degrees

&

Asinφ = 10 <=> A = 11.2 m

Thanks a ton for the help and the patience!

 

[Orodruin]

φ = 1.12 rad = 6.4 degrees

1.12 radians are not 6.4 degrees...

Also, a more direct approach to solve for A without computing the phase shift is
$$
A^2\cos^2\phi + A^2\sin^2\phi = A^2 = 100+25 = 125 \quad \Longrightarrow \quad A=\sqrt{125}\simeq 11.2
$$

 

[Const@ntine]

1.12 radians are not 6.4 degrees...

Also, a more direct approach to solve for A without computing the phase shift is
$$
A^2\cos^2\phi + A^2\sin^2\phi = A^2 = 100+25 = 125 \quad \Longrightarrow \quad A=\sqrt{125}\simeq 11.2
$$

I meant to write 63.4 but my keyboard's a bit busted (2,3 & the spacebar don't really work).