Homework Help: Can someone tell me if i'm using the right method? More wave fun!

1. Mar 25, 2006

mr_coffee

Hello everyone, my professor posted some powerpoint presentations and this seems to be the only equation that seems relatively close to figure out the answer. But i'm confused on the information given and not given.
Here is the problem:
A string along which waves can travel is 3.60 m long and has a mass of 280 g. The tension in the string is 33.0 N. What must be the frequency of traveling waves of amplitude 7.70 mm in order that the average power be 85.0 W?

Here is the formula it looks like i'm going ot use:
Average power transmited per unit length = .5*vKA^2;
v i'm assuming is the velocity of the string, now K i think is suppose to be the wave #.
A is suppose to be the amplitude. But i don't see how i'm going to find the wave # or v with the given. ANd plus they are giving me infromatino about tension which isn't even in this equation. Any ideas?

I also can find linear density by taking .28 kg/3.60m = .07778.
Thanks!

2. Mar 25, 2006

asrodan

You could try this equation

Average power per unit length = .5*A^2*v^2*(Tl)^.5

where A is amplitude, v is frequency, T is tension, and l is linear mass density

3. Mar 25, 2006

asrodan

always check to make sure units work out, if K is the wave number in the equation your prof gave you then the equation gives units

(1/s*1/m*m^2)=m/s, assuming there is no mistake in the equation (I think the frequency should be squared) then K would have to have units of kg/(s^2) to give a power.

The equation I gave you gives an answer that does have units of power.

4. Mar 25, 2006

mr_coffee

Thanks asrodan but it said it was wrong,
i did the folllowing:
Power = .5*A^2*v^2*(Tl)^.5
Solved for v and got::
v^2 = [2*(Power)]/(A^2*(TI)^(.5)
v = SQRT( [2*(85)] / (7.7E-3)^2 * (33*.0778)^(.5) )
v = 2143.28 Hz

5. Mar 25, 2006

asrodan

The answer I'm getting is 1337.71 Hz

If your using a program to calculate v, the way you have your equation (v=) written will put the (33*0.778)^(.5) in the numerator instead of the denominator.

Last edited: Mar 25, 2006
6. Mar 25, 2006

mr_coffee

O my bad, your right but I submitted: 1337.71 and it was still wrong. I think the formula might be off somewhere

7. Mar 25, 2006

asrodan

I dug out some of my textbooks to check the equation, and it is correct. However, the v should actually have been w (angular frequency) and the question seems to be asking for normal frequency in which case you need to divide by 2*Pi

8. Mar 25, 2006

mr_coffee

Thanks for checking that!
Well if i just found frequency instead of angualr and if v is frequency isn't the equation:
v = w/2*pi
so wouldn't I multiply 1337.71 by 2*pi rather then divide?

9. Mar 25, 2006

asrodan

The frequency you found from the equation I gave you was angular frequency.

10. Mar 25, 2006

mr_coffee

You are the man! ur getting cookies e-mailed to you asap.
thanks for the help!