Power of a Transverse wave question

[ultrapowerpie]

Homework Statement

Problem: http://img355.imageshack.us/my.php?image=a3physicsfa1.png

Homework Equations

http://hyperphysics.phy-astr.gsu.edu/Hbase/Waves/powstr.html

The Attempt at a Solution

Yeah, can someone help me exactly to find the velocity I need in order to plug it into the equation? I tried figuring out with the velocity using wavelength and all, but that didn't work. I think I need to sue that forumula, but without knowing T, I'm stumped right now.

Also, what exactly is the difference between the maximum speed of this wave, and the speed calculated from, let's say

V = sqrt (T/(M/L)) ?

 

[ultrapowerpie]

Please, need some help with this. >.>

 

[Redbelly98]

You can get the propagation velocity from wavelength and frequency. Can you show how you tried that, or just where you were stumped in your attempt?

 

[Doc Al]

Yeah, can someone help me exactly to find the velocity I need in order to plug it into the equation? I tried figuring out with the velocity using wavelength and all, but that didn't work.

The speed of the wave pulse is contained in the \sin (kx - \omega t) factor. How do k and ω relate to wavelength and frequency?

Read this: http://hyperphysics.phy-astr.gsu.edu/Hbase/waves/wavsol.html#c4"

 

[ultrapowerpie]

Ok, what exactly is this "propagation velocity". I've never heard of this term before, and I find it very... odd. Is it like initial velocity, average velocity, etc.

I know that k= 2pi/(wavelength)

I also know that f =w/2pi

I also know that v = (wavelength)*(frequency)

My question is, is that v the "propagation" velocity (the one in the above equation), whatever that means. Cause I believe I attempted finding the v that way, and got the answer wrong. This may have been just a math error on my part, but I wanted to double check.

I was also curious if that equation is actually need for this problem, or if it's just extraneous information.

Also, what's the difference between this "propagation velocity" and the maximum velocity of the wave?

Thanks for the help in advance!

 

[Redbelly98]

It sounds like you just had a math error somewhere, because

v = λ f

is the correct formula to use. If you say what you got for λ, f, and v, then we could tell you if that's right or wrong.

what's the difference between this "propagation velocity" and the maximum velocity of the wave?

Propagation velocity is the velocity at which the wave or pattern moves along the string. Check out the lower figure here:

http://www.isvr.soton.ac.uk/spcg/Tutorial/Tutorial/Tutorial_files/essaiwave.gif

The wave pattern just moves along, or propagates, to the right at the propagation velocity.

I'm not sure what "maximum velocity of the wave" refers to, since in this example the wave moves at a constant velocity. It might really mean the maximum velocity of a point on the string, i.e. the max velocity of the red dot in the above image. But that's just a guess on my part.

 

[ultrapowerpie]

Ok, I'll re-try the formula... Thanks, I probably did just make a mistake. I was just confused why they gave us a forumula.

There is a reason I'm asking about "Maximum" Velocity being different from this propagation velocity, it's another problem...

http://img377.imageshack.us/my.php?image=a4physicscn3.png

Now, problem 12, I figured out no problem.

But 13 asks for the "maximum transverse velocity", which is NOT the same as the answer to problem 12. This is why I brought this up, as I was now confused what this maximum transverse velocity is. Is that a fancy way of saying the angular frequency here, or is it some other value?

Thanks for clearing that up with the propagation velocity though. :D

 

[Redbelly98]

Ah, okay. The transverse velocity is the velocity of the string itself, for example the red dot in the figures of post #6. "Transverse" means at a right angle to the wave propagation direction; note that the red dot moves up-and-down, at a right angle to the right-going wave propagation direction.

 

[ultrapowerpie]

Ok, so, the difference here is:

The "propagation velocity" is the speed of the wave going in the x direction

and the "Transverse Wave Velocity" is the speed of the string going up and down (Y direction)

That right? Just wanted to make sure, in case I need to explain it to some class mates.

So, how do I find this transverse wave velocity with the info given? My book doesn't differentiate between the tranverse velocity and the propagation velocity (everything is just v >.>)

 

[Doc Al]

Ok, so, the difference here is:

The "propagation velocity" is the speed of the wave going in the x direction

OK.

and the "Transverse Wave Velocity" is the speed of the string going up and down (Y direction)

Careful with terminology. The question asks for "the maximum transverse speed of the particles in the string", which is what you describe. Don't call it the "transverse wave velocity"--that's confusing. (If someone asked me for the "transverse wave velocity", I'd assume that they meant to ask for the velocity of the transverse wave, which is just v.)

So, how do I find this transverse wave velocity with the info given? My book doesn't differentiate between the tranverse velocity and the propagation velocity (everything is just v >.>)

Start with the wave form, which is something like: y = A sin(kx - ωt). That's the y (transverse) coordinate as a function of time and position. Hint: The transverse speed is dy/dt.

 

[ultrapowerpie]

See, that's the thing. My book uses cos as the initial, but other websites use sin. There is a different between the two functions, despite my professor's claim that they can be both used. I don't believe they can be used interchangably, but that's just me.

Thank you for correcting my terminology. I figured a derivative was needed, but when I tried it earlier, I got a number much lower then the normal velocity. Now that I know they're different things, I can retry the problem and hopefully get the right result, I'll post here with the fruits of my efforts.

Thank you again. :D

 

[Doc Al]

See, that's the thing. My book uses cos as the initial, but other websites use sin. There is a different between the two functions, despite my professor's claim that they can be both used. I don't believe they can be used interchangably, but that's just me.

Your professor is correct; either sine or cosine can be used. Sure they are different functions, but only by a phase factor. (Try it and see!)

 

[DEW]

Yep, that's right. Just wanted to join in - too late as usual !

 

[ultrapowerpie]

Ok, I'm having a problem with 11 (the one in my first post).

I calculated my V = 665.62

I'm certain this value is right.

However, there's a problem. The answer I get by plugging in the data, is .014. This answer is wrong.

I have a stong suspicion that the amplitude for this equation is NOT A0, but another number. As the w is given, and so is the density, the only thing that looks odd is the amplitude.

So, here's the question: They gave me an x value. Obviously, I need to use it somehow. Is the amplitude A0*e^(-bx)? Cause I tried using that for the amplitude, and got the problem wrong.

 

[ultrapowerpie]

Scratch the question for 13 I asked earlier in this post, I just realized there was a much simpler formula-

Vmax= w*A

And that is right. Just need to work on problem 11

 

[Doc Al]

Ok, I'm having a problem with 11 (the one in my first post).

I calculated my V = 665.62

I'm certain this value is right.

How did you calculate this value?

 

[ultrapowerpie]

I took the K value, and did 2pi/k = 14.61

Then, I did f = w/2pi = 4.615...

Whoops, I just made a mistake in my math apparently, hold on...

 

[ultrapowerpie]

Is the velocity 67.44? And what about my amplitude theory I made before?

 

[Doc Al]

Is the velocity 67.44?

That sounds better.

And what about my amplitude theory I made before?

I agree with what you said earlier:

Is the amplitude A0*e^(-bx)?

That's what I would use. The formula you're plugging into is for a sinusoidal wave; your pulse can be considered a sine wave modulated by that amplitude.

 

[ultrapowerpie]

Thank you, I will try that, give me a bit...

 

[ultrapowerpie]

Got it right, thanks! :D