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-   -   My tricky wave problem. ehh its probably not tricky. (http://www.physicsforums.com/showthread.php?t=213146)

 bobbo7410 Feb4-08 07:40 PM

My tricky wave problem. ehh its probably not tricky.

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 Feb4-08 11:57 PM

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 Feb5-08 12:09 AM

What happens when you substitute $$\omega = 2\pi f = \frac{2\pi v}{\lambda}$$

 bobbo7410 Feb5-08 12:33 AM

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.

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

 Dick Feb5-08 12:45 AM

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 Feb5-08 12:49 AM

Quote:
 Quote by bobbo7410 (Post 1597301) 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.
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 Feb5-08 12:53 AM

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 Feb5-08 12:57 AM

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 Feb5-08 01:03 AM

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 Feb5-08 01:10 AM

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 Feb5-08 01:15 AM

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 Feb5-08 01:24 AM

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

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

 Dick Feb5-08 01:26 AM

Units checking goes for you as well, mezarashi!

 mezarashi Feb5-08 01:34 AM

Quote:
 Quote by Dick (Post 1597344) Units checking goes for you as well, mezarashi!
;D

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

 bobbo7410 Feb5-08 01:40 AM

yup! I redid the equation just like that and got it correct, regardless if I didn't get any points. :cry: 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. :confused:

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

 mezarashi Feb5-08 01:57 AM

Quote:
 Quote by bobbo7410 (Post 1597353) ????? 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. :confused:
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.

Quote:
 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.

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