Fundamental frequency of violin string

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The discussion revolves around calculating the fundamental frequency of a violin string that is shortened by pressing it against the fingerboard. The original string length is 330 mm with a fundamental frequency of 659 Hz, and when pressed at 60 mm from the end, the new length becomes 270 mm. Participants clarify that the original frequency is necessary to determine the new frequency using the formula f = nv/2L, where L is the new length. The importance of understanding that the velocity of the wave on the string must be derived from the original frequency is emphasized. Ultimately, the new fundamental frequency is calculated to be 805 Hz.
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Homework Statement



A vibrating string on a violin is 330 mm long and has a fundamental frequency of 659 Hz. What is its fundamental frequency when the string is pressed against the fingerboard at a point 60 mm from its end?

*The answer is 805 Hz

Homework Equations


f = \overline{}nv/2L
wavelength = v/f

The Attempt at a Solution


I don't understand what to do with the 60mm. It splits the string into 2 unequal parts, which isn't a harmonic I recognize.
 
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I don't understand what to do with the 60mm. It splits the string into 2 unequal parts, which isn't a harmonic I recognize.
That's just telling you how much shorter the string is now. You are ignoring the 60mm part that is pinched off, and considering the length that remains. This is the length you are trying to find the fundamental frequency for.
 
hage567 said:
That's just telling you how much shorter the string is now. You are ignoring the 60mm part that is pinched off, and considering the length that remains. This is the length you are trying to find the fundamental frequency for.

Oh, so there's no node there? So (330-60) mm (=270) is the new length, L, of the string. What do you do with the original frequency they gave you?
By the way, this is a string with a fixed node on both ends, right?
 
Oh, so there's no node there? So (330-60) mm (=270) is the new length, L, of the string.
By the way, this is a string with a fixed node on both ends, right?
Yes, that's correct.
What do you do with the original frequency they gave you?

You will need it to figure out the fundamental frequency of the 270mm length of the string. :smile:

You have the equations you need to solve this. You just need to find a way to relate the first string to the second string.
 
hage567 said:
Yes, that's correct.
You just need to find a way to relate the first string to the second string.

That's what I don't understand--how are the two connected? They're not harmonics. I keep trying to use the formula f = nv/2L, but that doesn't work and it doesn't use the original frequency.
Intuitively, it seems that shortening the length of an already-vibrating string would increase the frequency, but I don't know how to derive this mathematically.
 
Oh ok, I got it--I assumed that the velocity was 343 m/s, but you have to use the original frequency to calculate it.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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