Fundamental Frequency of a string

  • Thread starter nn3568
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  • #1
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Homework Statement


The length of a string is 1440 cm. The
string is held fixed at each end. The string
vibrates in eight sections; i.e., the string has
eight antinodes, and the string vibrates at
150 Hz.

What is the fundamental frequency? Answer
in units of Hz.


Homework Equations



f = nv / 2L frequency = (# antinodes)(velocity) / (2 * length)

The Attempt at a Solution



I already found the wavelength which is 3.6 m. I got that by substituting nv/2L for f into the equation v=λf (velocity = wavelength*frequency). Then I followed the above frequency formula:

v=λf
v=(3.6 m)(150 Hz) = 540 m/s

f = nv / 2L
f = (8)(540 m/s) / (2*14.4 m) = 150 Hz

My online HW said it was wrong. So I tried again.

f = (2)(540 m/s) / (2*14.4 m) = 37.5 Hz

I used two because it said that fundamental frequency is the lowest frequency and you can use the first harmonic or something. Didn't really understand it. But the two is because the 1st harmonic has 2 antinodes.

I am so confused and I have no clue what I am doing! Please help <:S
 
Last edited:

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
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If there are 8 antinodes then there are 4 wavelengths represented on the string.

But I think 8 antinodes means it's the 8'th harmonic of the string.

So isn't 1/8 of 150 hz then the first harmonic and isn't that the frequency you use with 2*L to determine v?
 
  • #3
LowlyPion
Homework Helper
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Or do you not need to find the v?

Only the f1 fundamental frequency?
 
  • #4
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i already found v. i just dont understand the fundamental frequency part.
 
  • #5
LowlyPion
Homework Helper
3,090
4
i already found v. i just dont understand the fundamental frequency part.
8 antinodes means 8th harmonic = 8 times the fundamental frequency (first harmonic).
 
  • #6
LowlyPion
Homework Helper
3,090
4
Here's a link:
http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/sound/u11l4d.html [Broken]
 
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