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

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Homework Help Overview

The discussion revolves around a physics problem involving wave mechanics, specifically related to the frequency of traveling waves on a string given certain parameters such as tension, mass, and amplitude. Participants are exploring the application of formulas related to average power transmitted through the string.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply a formula for average power but expresses confusion regarding the variables involved, particularly the wave number and velocity. Other participants suggest alternative equations and provide insights on unit consistency and the relationship between angular frequency and normal frequency.

Discussion Status

Participants are actively engaging with the problem, offering different equations and checking each other's calculations. There is an ongoing exploration of the correct interpretation of variables and formulas, with some participants questioning the accuracy of the original approach and suggesting corrections.

Contextual Notes

There is a noted uncertainty regarding the use of tension in the equations and the distinction between angular frequency and normal frequency. The original poster has expressed difficulty in reconciling the given information with the equations being used.

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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!
 
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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
 
Almost forgot, some general advice

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.
 
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
 
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:
O my bad, your right but I submitted: 1337.71 and it was still wrong. I think the formula might be off somewhere
 
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
 
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?
 
The frequency you found from the equation I gave you was angular frequency.
 
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You are the man! ur getting cookies e-mailed to you asap.
thanks for the help!
 

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