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My tricky wave problem. ehh its probably not tricky.

by bobbo7410
Tags: tricky, wave
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bobbo7410
#1
Feb4-08, 06:40 PM
P: 36
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 : ) **
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bobbo7410
#2
Feb4-08, 10:57 PM
P: 36
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
#3
Feb4-08, 11:09 PM
HW Helper
P: 660
What happens when you substitute [tex]\omega = 2\pi f = \frac{2\pi v}{\lambda}[/tex]

bobbo7410
#4
Feb4-08, 11:33 PM
P: 36
My tricky wave problem. ehh its probably not tricky.

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
#5
Feb4-08, 11:45 PM
Sci Advisor
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P: 25,228
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
#6
Feb4-08, 11:49 PM
HW Helper
P: 660
Quote Quote by bobbo7410 View Post
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
#7
Feb4-08, 11:53 PM
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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
#8
Feb4-08, 11:57 PM
P: 36
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
#9
Feb5-08, 12:03 AM
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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
#10
Feb5-08, 12:10 AM
P: 36
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
#11
Feb5-08, 12:15 AM
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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
#12
Feb5-08, 12:24 AM
HW Helper
P: 660
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
#13
Feb5-08, 12:26 AM
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Units checking goes for you as well, mezarashi!
mezarashi
#14
Feb5-08, 12:34 AM
HW Helper
P: 660
Quote Quote by Dick View Post
Units checking goes for you as well, mezarashi!
;D

And now I feel guilty he couldn't complete his question on time >_<
bobbo7410
#15
Feb5-08, 12:40 AM
P: 36
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
#16
Feb5-08, 12:57 AM
HW Helper
P: 660
Quote Quote by bobbo7410 View Post
????? 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.
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.

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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