Sinusodial wave motion and reflection

[talking_chicken]

Homework Statement

Okay, I've got a question that's been bugging me for the longest time.

I've got a string attached to a wall at one end (the other end is free to move, though) and it has a sinusoidal wave traveling to the right and hits the wall at x = L and reflects. I'm supposed to prove that the sum of the leftward and rightward waves are

yR(x,t) + yL(x,t) = 2Asin(k(x - L))cos(kvt + phi)

which I did (unless there's a mistake somewhere), which makes a nice standing wave. but now they want me to solve for phi (the phase shift) in terms of k, L, and v. No numbers are given, it's all variables.

Homework Equations

yR(x,t) + yL(x,t) = 2Asin(k(x - L))cos(kvt + phi) <--- equation of the standing wave
yR(x,t) = Asin(k((x - L) - vt) - phi) <--- rightward wave equation (I hope)
yL(x,t) = Asin(k((x - L) + vt) + phi) <--- leftward wave equation (I hope)
k = 2pi / lambda
v = lambda / T
k = omega / v

k = wave number
v = velocity of wave
lambda = wavelength
phi = phase shift
y = height of section of wave at time t
T = period

The Attempt at a Solution

Okay, I know that at t = 0, parts of the wave actually hit the maximum height of the wave (2A). I found out that the parts of the rope hit this height are at x = L - n*lambda/4 = L - n*pi/2k, but putting this into the formula just gets rid of L completely, which is probably something I don't want to do. I did find out that phi = -kvt (set x to L - n*2pi/k and solve for phi), but I need L, not t.

I tried substituting x = 0 and t = 0 right away, but the problem is I don't know any initial values and just get y = 2Asin(-kL)cos(phi). This leads to cos(phi) = y(0,0)/2Asin(-kL), which I don't know helps or not. Besides, setting t=0 gets rid of the v term, which I think I need.

Differentiation gets me nowhere, so far. I tried to differentiate before adding and adding before differentiating and seeing if I can get an equation out of that, to no avail. Doing dy/dt brings out the v, but I don't know what to do with that (finding the max height means setting dy/dt to 0, which just gets rid of the constant coefficients, like v).

I'm really stuck and have been looking at this for about two days. Help!

 

[anathema]

Thank you for your question. It seems like you have already made some good progress in solving for the phase shift phi. To continue, let's start by looking at the equation for the standing wave:

yR(x,t) + yL(x,t) = 2Asin(k(x - L))cos(kvt + phi)

We can rewrite this equation as:

yR(x,t) + yL(x,t) = 2Asin(kx - kL)cos(kvt + phi)

Now, let's consider the two individual wave equations:

yR(x,t) = Asin(k((x - L) - vt) - phi)
yL(x,t) = Asin(k((x - L) + vt) + phi)

If we add these two equations together, we get:

yR(x,t) + yL(x,t) = Asin(k((x - L) - vt) - phi) + Asin(k((x - L) + vt) + phi)

Using the trigonometric identity sin(a + b) = sin(a)cos(b) + cos(a)sin(b), we can rewrite this as:

yR(x,t) + yL(x,t) = Asin(k(x - L))(cos(vt)cos(phi) - sin(vt)sin(phi)) + Asin(k(x - L))(cos(vt)cos(phi) + sin(vt)sin(phi))

Simplifying this further, we get:

yR(x,t) + yL(x,t) = Asin(k(x - L))(2cos(vt)cos(phi))

Now, we can compare this to our original equation for the standing wave:

yR(x,t) + yL(x,t) = 2Asin(k(x - L))cos(kvt + phi)

We can see that the two equations are equal if:

2cos(vt)cos(phi) = cos(kvt + phi)

Using the trigonometric identity cos(a + b) = cos(a)cos(b) - sin(a)sin(b), we can rewrite this as:

2cos(vt)cos(phi) = cos(kvt)cos(phi) - sin(kvt)sin(phi)

Comparing this to our original equation, we can see that:

2cos(phi) = cos(phi)

This means that cos(phi) = 1/2. Solving for phi, we get:

phi =

 

[m2003]

Hello there, it seems like you have a pretty good understanding of sinusoidal wave motion and reflection. To solve for phi, the phase shift, in terms of k, L, and v, you can use the equations you have already derived and the concept of superposition.

First, let's define phi as the phase shift between the leftward and rightward waves. This means that when the waves meet at the wall at x = L, the leftward wave will have a phase shift of phi compared to the rightward wave. We can express this as:

yL(x,t) = Asin(k((x - L) + vt) + phi)

yR(x,t) = Asin(k((x - L) - vt) - phi)

Now, when these two waves meet at x = L, they will add up to form a standing wave. This means that the sum of these two equations should equal the equation for the standing wave that you have already derived. So we can write:

yL(x,t) + yR(x,t) = 2Asin(k(x - L))cos(kvt + phi)

Substituting in the equations we have for yL(x,t) and yR(x,t), we get:

Asin(k((x - L) + vt) + phi) + Asin(k((x - L) - vt) - phi) = 2Asin(k(x - L))cos(kvt + phi)

Expanding the left side using the trigonometric identity sin(a+b) + sin(a-b) = 2sin(a)cos(b), we get:

2Asin(k(x - L))cos(vt)cos(phi) + 2Acos(k(x - L))sin(vt)sin(phi) = 2Asin(k(x - L))cos(kvt + phi)

Since the two sides are equal, we can equate the coefficients of sin(k(x - L)) and cos(k(x - L)). This gives us two equations:

2Acos(vt)cos(phi) = 2Asin(k(x - L))

2Acos(k(x - L))sin(vt) = 2Asin(k(x - L))cos(phi)

Dividing the first equation by the second, we get:

tan(phi) = cos(vt)/sin(vt)

Using the identity tan(phi) = sin(phi)/cos(phi), we can rewrite this as