My tricky wave problem. ehh its probably not tricky.

bobbo7410

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

a wave on a cable, mass per unit length(mu) 1.70 kg/m, is created by a machine providing 250(p) W of power. It creates waves of wavelength(lambda) 2.90 m and an amplitude(A) 2.90 m. What's the speed of these waves?

2. Relevant equations

v = lamda/frequency
k = 2pi/lamda
w = 2pi/t = 2pi*frequency
etc

3. The attempt at a solution

i am given mu , p , a , lamba as well i can find k

Since they give power and that is a main equation in the chapter, I would use the equation: p = (1/2)mu * w^2 * A^2 * v and solve for v. simple.

butttt I simply cannot find the correct equation to find "w" the angular frequency. Every equation I've looked at it seems I'm just short 1 variable.

if I could find T(period) or w(angular frequency) or f(frequency) I could figure out the solution.


This is my first time checking out these forums. Any input would be greatly appreciated!

** ill be online refreshing constantly for the next couple hours : ) **

bobbo7410

sry to "bump" if its not allowed.

I considered setting 2 velocity equations equal to eachother and solving for the variables.

[ f * wavelength = square root ( T / mu ) ]

If I could solve for f or T I could get the final velocity. Yet that proposes 2 unknown variables so I believe I am still stuck.

The assignment is due in about an hour so I'll continue to check until then! Thanks everyone!

mezarashi

What happens when you substitute [tex]\omega = 2\pi f = \frac{2\pi v}{\lambda}[/tex]

bobbo7410

Thanks for the help! I thought of something similar to that to isolate v. I substituted your suggestion into the original equation. It seemed kind of long and drawn out as if there must be a simpler way. I doubt I have any simple math errors.

The answer turned up to be incorrect.

I attached the full pic of the work either way.

http://www.uploadyourimages.com/view...0204082330.jpg


I have 27 min left. oooo this is getting close.

Dick

If P=(1/2)*mu*w^2*A^2*v, and w=(2pi)/f and v=lambda/f, it would seem that w=(2pi*v)/lambda. it seems to me you know everything in the P equation except for v. Can't you just solve for it?

mezarashi

If you were to have taken my suggestion, and done the substitution, you will find that you are only solving a simple algebraic equation of the 1st degree. That is, ax + b = c, solve for x where a,b,c are constants. With the best of intentions, I suggest you take up some tutoring on algebra if you honestly find solving such a problem difficult.

Dick

w has the units of 1/sec. (2pi*lambda/v) has the units of sec. (2pi*v)/lambda has the units of 1/sec. Which do you believe? Keeping track of units is G*d's gift to man. Uh, and women.

bobbo7410

so I probably should have used (2pi*v)/lambda

and I suppose im just dumb then... I cannot figure out the simple way to solve for the answer than how I did it. I simply plugged in for w the equation and solved the rest.

meza please elaborate more if you could.

Dick

You get v^3 on one side and a bunch of numbers on the the other. Then just take a cube root. Keep track of the units and make sure your answer comes out in m/sec. Otherwise, it's garbage.

bobbo7410

yea as I'm working the equation, if I would have used [(2pi*v)/lambda] instead of [(2pi*lambda)/v] I would have gotten it correct. I get v^3 equal to a #m/s.

times up, oh well. but thanks dick!

but in doing that I just did the same as before, I simply plugged that as w and solved P=(1/2)*mu*w^2*A^2*v

yet meza suggests I made that insanely difficult and that I need to re-look into basic algebra. (albeit he suggested an incorrect equation) Can one of you maybe elaborate on what I am totally overlooking then in that aspect?

Thanks dick and meza though for all your help!

Dick

You got the v and lambda inverted in the w solution. No big deal, it happens to everyone. But learn how to check your work by tracking units. You can find a lot of mechanical errors that way.

mezarashi

My original equation was incorrect. Sorry about that. I had the v and lambda exchanged. In anycase, the equation as given is:

[tex]P = \frac{1}{2}v \mu \omega^2 A^2[/tex]
You know P, you know mu, you know A. Omega can be expanded into:
[tex]\omega = 2\pi f = \frac{2\pi v}{\lambda}[/tex]
which means the only unknown variable here is v:
[tex]P = \frac{1}{2}v \mu 4 \pi^2 v^2 A^2[/tex]

Dick

Units checking goes for you as well, mezarashi!

mezarashi

;D

And now I feel guilty he couldn't complete his question on time >_<

bobbo7410

yup! I redid the equation just like that and got it correct, regardless if I didn't get any points. I did the same work as the first time, just switched that equation.

http://img149.imageshack.us/img149/8...5080033qn2.jpg

But I'm still hung up on,
"If you were to have taken my suggestion, and done the substitution, you will find that you are only solving a simple algebraic equation of the 1st degree. That is, ax + b = c, solve for x where a,b,c are constants. With the best of intentions, I suggest you take up some tutoring on algebra if you honestly find solving such a problem difficult."

????? what was wrong with the way I did it? Using that exact formula, I can't see how I was so wrong in what I did.


ha dont worry about it mez, I'm just happy someone even offered to help me out!

mezarashi

No, I had meant that for your previous comment, below saying "long and drawn out". I was supposing all physics problems are "long and drawn out" =P. Now looking at your work, your algebra is fine ^^. Sorry if it didn't really get across right.