# Waves and Wave equation

1. Jan 23, 2013

### Stealth849

1. The problem statement, all variables and given/known data

Hello all,

stuck on a question involving a formula for a wave that doesn't make much sense to me.

Assuming that a wave on a string is represented by:

y(x,t) = y_i*sin((2∏/λ)(vt-x))

Where y is transverse displacement at time t of the piece of string at x. The other symbols have their "usual meaning". Find the velocity and acceleration of the small piece of string at x = 10m, as a function of time.

Making use of the fact that the piece of string satisfies Newton's second law, show that the piece of string is acted on by a Hooke's Law force.

2. Relevant equations

y(x,t) = y_i*sin((2∏/λ)(vt-x))

y(x,t) = A*sin(kx - ωt)

k = 2∏/λ

3. The attempt at a solution

I'm a bit confused by the format of the formula...

I know that to find velocity and acceleration, I can take the derivative of the equation once for velocity and again for acceleration. Looking at the set up, it looks like if we replace 2∏/λ with k and expand, we see in the brackets

(kvt - kx)

where kv could equal ω?

I don't know if this is at all valid. I also don't understand why the variable "v" is used in the equation, but from the question i can assume that it is analogous to omega ω.

So

y(10,t) = y_i*sin((2∏/λ)(vt-10))

v(10,t) = y_i*cos((2∏/λ)(vt-10))*???

I'm having trouble knowing how to take the derivative of this when I don't know λ and v.
Unless they are constants and I can ignore them?

Any help will be appreciated... Thanks.

2. Jan 23, 2013

### cepheid

Staff Emeritus
The usual form for a wave is something like sin(kx - ωt). The phase is given by kx - ωt, so that if x is constant (i.e. you're sitting at a particular point in space as the wave passes by you) the variation of wave intensity with time is sinusoidal with angular frequency ω. When t = T, where T is one period of the oscillation ($T \equiv 1/f = 2\pi /\omega$), then ωt = ωT = 2π. The phase increments by 2π as expected after one cycle. Note: f is the frequency of the wave, and ω = 2πf is the angular frequency.

You can rewrite kx - ωt as k[x - (ω/k)t] $$= k\left(x - {2\pi f}\frac{\lambda}{2\pi}t\right)$$ $$= k(x -f\lambda t)$$

Of course, there is just the standard relation that wave speed = frequency*wavelength: $v = f\lambda$ so this becomes k(x - vt). This form of the wave function is actually kind of intuitive, because it says that at time t, the amplitude of the wave at a distance vt ahead of a point x is the same as what the amplitude was at point x back at time 0.

EDIT: And yes, of course f and λ (and hence v) are constant for a given wave. Don't confuse "v" (the wave propagation speed, in the x direction) with dy/dt, the speed of a point oscillating in the y direction as the wave passes by, which you are asked to compute.

Last edited: Jan 23, 2013
3. Jan 24, 2013

### tms

$v$ is just the speed of the wave, as distinct from the transverse speed of the bits of string.
For a given wave, they are constant.

4. Jan 24, 2013

### Stealth849

Okay..

So considering that λ and v are constants, the equations should look, hopefully, something like:

y(10,t) = y0*sin[k(vt-10)]

v(10,t) = y0*cos[k(vt-10)]*k

a(10,t) = -k2y0*sin[k(vt-10)]

I assume this is as far as I can go? I am given nothing else but distance x = 10m, so I cannot find the amplitude, or any of the constants, can I?

I also don't understand what the second part of the question is asking, when it asks to show the string is acted on by a Hooke's Law force knowing that the string follows Newton's second law.

What do the two equations have in common, and how can I show that the string is modeled by F = -kx, from F = ma?

5. Jan 24, 2013

### tms

Check your derivatives again. Remember the chain rule.

As for the second part, you have just calculated the acceleration, and you already know the displacement. Just set $ma = -kx$, or $y$ in this case. And be sure to keep the two $k$s straight; use a different letter for one of them, perhaps.

6. Jan 24, 2013

### Stealth849

Is the chain rule not just multiplying the initial derivative by the derivative of what is in the brackets of the sin function?

where k(vt-10) is a product rule?

if k is a constant, derivative is zero, derivative of (vt - 10) should be v (which I forgot to put in... argh)

So equations should be:

y(10,t) = y0*sin[k(vt-10)]

v(10,t) = y0*cos[k(vt-10)]*k*v

a(10,t) = -k2v2y0*sin[k(vt-10)]

and if a(10,t) = -k2v2y0*sin[k(vt-10)]

I set that equation, multiplied by m, to y? y being the initial displacement equation..?

Also what do you mean by keeping the k's straight, are they not the same value as a constant..?

thanks

7. Jan 24, 2013

### tms

Easier to just multiply it out in this case: $kvt - 10k$.
That's it.
Start with $F = ma$, with $a$ as derived above. Then look at that equation and compare it to your equation for $y$. The answer should jump out at you.
The $k$ in Hooke's law and the $k$ in the equation for the wave are entirely different entities that just happen to use the same letter.

8. Jan 24, 2013

### Stealth849

Ah, thought you were referring to the k's in the derived equation. Sorry!

Anyway, if

F = ma = -kx

Would x in this case be y? and thus y0sin[k(vt-10)]

so ultimately we see

-mk2v2y0*sin[k(vt-x)] = -Ky0sin[k(vt-x)]

where uppercase K is spring constant, the amplitudes cancel, the sin functions cancel, and were left with

mk2v2 = K

But I don't see how that actually says anything..? Nothing jumped out at me. :(

9. Jan 24, 2013

### tms

You've got all the parts, you just have to put them together correctly.

Think of the form of Hooke's law: the force equals -1 times a constant times the displacement. Now take your force: $F = ma = m$ times the acceleration you've derived. How can you put that in the form of Hooke's law? Remember that you also have an expression for the displacement, that is $y$.

Last edited: Jan 24, 2013
10. Jan 24, 2013

### Stealth849

Well, if we replace the y0sin[k(vt-10)] portion of the F = ma equation, that would give

F = -mk2v2y

This somewhat models Hooke's law, if we can consider m, k, and v to be all constants and serve as the constant in the Hooke's law equation.

Is this more on the right track..?

11. Jan 24, 2013

### tms

You've got it. $m$, $k$, and $v$ are all constants in this situation.