Calculating harmonics of a copper wire

AI Thread Summary
The discussion focuses on calculating the wavelength of the third harmonic for a copper wire with a specific length and tension caused by hanging weights. The initial calculation yields a wavelength of 0.75m using the formula λ = 2L/n. However, the tension change when replacing a 200N ball with a 1000N ball affects the wire's length, which is crucial for accurate wavelength determination. The ambiguity in whether the given length of 1.13m is under tension or tension-free is noted as significant for solving the problem. Young's modulus is suggested as a method to relate tension to the wire's length for further calculations.
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Homework Statement


A vertical, 1.13m length of 18-gauge (diameter of 1.024mm) copper wire has a 200.0N ball hanging from it.

What is the wavelength of the third harmonic for this wire?

A 1000- ball now replaces the original ball. What is the change in the wavelength of the third harmonic caused by replacing the light ball with the heavy one? (Hint: See Table 11.1 in the textbook for Young's modulus.)


Homework Equations


[lambda=(2L)/n

L=length of string
n=harmonic number


[h2]The Attempt at a Solution[/h2]

1. lambda=(2L)/n
so =2/3*1.13=0.75m
But the next question implies that it's related to tension. How?

 
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gboff21 said:

Homework Statement


A vertical, 1.13m length of 18-gauge (diameter of 1.024mm) copper wire has a 200.0N ball hanging from it.

What is the wavelength of the third harmonic for this wire?

A 1000- ball now replaces the original ball. What is the change in the wavelength of the third harmonic caused by replacing the light ball with the heavy one? (Hint: See Table 11.1 in the textbook for Young's modulus.)


Homework Equations


[lambda=(2L)/n

L=length of string
n=harmonic number


[h2]The Attempt at a Solution[/h2]

1. lambda=(2L)/n
so =2/3*1.13=0.75m
But the next question implies that it's related to tension. How?
The formula

λ = 2L/n (where n = 1 corresponds to the fundamental wavelength)​

applies the same regardless of the tension.

But what's different when you replace the 200 N ball with a 1000 N ball is the length of the copper wire changes a little bit. In other words, L is not the same in each case.

(The frequency of the wave changes a lot. The wavelength only changes a little, but it does change because L changes a little.)

The thing which I question is whether the problem statement claims that length is 1.13 m when tension free, or if it is 1.13 m when the 200 N ball is already hanging. It sort of sounds to me like the wire is 1.13 m when the 200 N ball is already hanging. That's my guess anyway. But it's only a guess. The problem statement is kind of ambiguous regarding this. Perhaps your coursework has a figure or something that would make it more obvious which one it is. You'll have to determine which one, because it makes a difference.

Whatever the case, you can use Young's modulus to find the relationship between the tension on the wire and its length. Once you know the wire's length repeat what you did above to find the wavelength of the third harmonic.
 
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