# My tricky wave problem. ehh its probably not tricky.

• bobbo7410
In summary: Physics forum! But really, where was I wrong?In summary, we are trying to find the speed of waves on a cable with a mass per unit length of 1.70 kg/m, created by a machine providing 250 W of power, with a wavelength of 2.90 m and an amplitude of 2.90 m. We use the equation P = (1/2)mu*w^2*A^2*v and solve for v, by substituting w = (2pi*v)/lambda. This simplifies the equation to P = (1/2)v*mu*4*pi^2*v^2*A^2, and solving for v yields the speed of the
bobbo7410

## Homework Statement

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?

## Homework Equations

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

## 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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sry to "bump" if its not allowed.

I considered setting 2 velocity equations equal to each other 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!

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

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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/4985330204082330.jpgI have 27 min left. oooo this is getting close.

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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?

bobbo7410 said:
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.

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.

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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.

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

and I suppose I am 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.

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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.

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!

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.

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:
$$P = \frac{1}{2}v \mu 4 \pi^2 v^2 A^2$$

Units checking goes for you as well, mezarashi!

Dick said:
Units checking goes for you as well, mezarashi!

;D

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

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/img149/8354/0205080033qn2.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 don't worry about it mez, I'm just happy someone even offered to help me out!

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bobbo7410 said:
? 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.

## 1. What is a tricky wave problem?

A tricky wave problem is a scientific puzzle that involves understanding the behavior and properties of waves. It may involve complex calculations or experimental observations to solve.

## 2. How do I approach solving a tricky wave problem?

First, identify the type of wave involved and its properties, such as wavelength, frequency, and amplitude. Then, use relevant equations and principles to analyze the problem and come up with a solution. It may also be helpful to draw diagrams or create simulations to visualize the problem.

## 3. What are some common mistakes when solving tricky wave problems?

One common mistake is not considering all the relevant variables and equations. Another is not understanding the underlying principles and concepts, leading to incorrect assumptions. It is also important to check units and conversions, as well as accuracy in calculations.

## 4. Can you provide an example of a tricky wave problem?

Sure, an example of a tricky wave problem could be determining the frequency of a sound wave based on its wavelength and the speed of sound in a particular medium. This would require understanding the relationship between frequency and wavelength, as well as the properties of sound waves in different mediums.

## 5. How can I improve my skills in solving tricky wave problems?

Practice and repetition are key to improving problem-solving skills in any scientific discipline. It is also helpful to seek guidance from experienced scientists or educators, as well as utilizing online resources and textbooks to deepen your understanding of wave properties and principles.

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