What is the Wave Equation for a String?

[Stealth849]

Homework Statement

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.

Homework Equations

y(x,t) = y_i*sin((2∏/λ)(vt-x))

y(x,t) = A*sin(kx - ωt)

k = 2∏/λ

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.

 

[cepheid]

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.

 

[tms]

I don't know if this is at all valid. I also don't understand why the variable "v" is used in the equation,

v is just the speed of the wave, as distinct from the transverse speed of the bits of string.

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?

For a given wave, they are constant.

 

[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?

 

[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 ks straight; use a different letter for one of them, perhaps.

 

[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

 

[tms]

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?

Easier to just multiply it out in this case: kvt - 10k.

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

That's it.

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..?

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.

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

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.

 

[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. :(

 

[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.

 

[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..?

 

[tms]

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