# Resonant Frequency of a tuning fork

• KYPOWERLIFTER
In summary: Clearly, I was being stupid.In summary, the lowest resonant frequency of a tuning fork with each tine being 18 cm long and a speed of wave of 420 m/s is 583 Hz. The first overtone is 1750 Hz, and the second overtone is 2917 Hz. This is calculated using the formula F=Velocity/Wavelength and considering the tine length as a node-antinode relationship, resulting in a node at one end and an antinode at the other end. The equations used are l=1/4(Lambda), l=3/4(Lambda), and l=5/4(Lambda).
KYPOWERLIFTER

## Homework Statement

Name the two lowest resonant frequencies of a tuning fork; each tine is 18 cm long. Speed of wave is 420m/s... ignoring metal thickness.

## Homework Equations

F= Velocity/Wavelength; l= wavelength/2, l= wavelength, l= 3/2wavelength, l= 2wavelength, l= 5/2wavelength

## The Attempt at a Solution

fundamental = 420m/s / .72m = 583.3 Hz
first overtone = 420m/s / .36m = 1167 Hz
second overtone = 420m/s .24m = 1750 Hz

The given answer is 583 Hz
1750 Hz
2917 Hz

I would have expected to use the formulas wavelength/2, wavelength, 3/2wavelength. They have used, wavelength/2, 3/2wavelength, and 5/2wavelength. Is this because of the antinode-node boundary for each tine? I'm confused as to why they make the frequency jumps that they do. Very elementary physics and I am no student. So, I have no one to ask.

KYPOWERLIFTER said:

## Homework Equations

F= Velocity/Wavelength; l= wavelength/2, l= wavelength, l= 3/2wavelength, l= 2wavelength, l= 5/2wavelength

These equations aren't right. If $L=\lambda/2$ then it means that there is a node at both ends. But one end of the tine is free. So you must have a node at one end and an antinode at the other. Does that help?

Tom Mattson said:
These equations aren't right. If $L=\lambda/2$ then it means that there is a node at both ends. But one end of the tine is free. So you must have a node at one end and an antinode at the other. Does that help?

Thanks for the reply... The equations are correct in the case of an antinode at each end. I took it that the book's writer must be working it that way, even though the fork has a node at the handle and two antinodes at the ends.

If he is basing it as if it's a node and an antinode then it would be $L=\lambda/4$, $L=\3/4lambda$, etc. This would be for two different wavea, though? One in each tine? That is to say, the 'L' would be 18cm? Twice? Even so, his method ofworking the problem does not compute that way, especially in light of:

I do not understand why he worked the problem by what seems to me is 'skipping' every other overtone?

Is my question clearly stated? Thanks, again

KYPOWERLIFTER said:
Thanks for the reply... The equations are correct in the case of an antinode at each end.

But you don't have an antinode at each end. One end of the tine is fixed.

Yup. Using the l=1/4(Lambda), l=3/4(Lambda), l=5/4(Lambda). Considering the system as a node-antinode relationship (which it clearly is), and thus employing .18m as the tine length, viola!

Thanks

## 1. What is the resonant frequency of a tuning fork?

The resonant frequency of a tuning fork is the specific frequency at which the tuning fork vibrates and produces its loudest sound. This frequency is determined by the physical characteristics of the tuning fork, such as its length, width, and material.

## 2. How is the resonant frequency of a tuning fork measured?

The resonant frequency of a tuning fork can be measured using a frequency counter or a oscilloscope. The tuning fork is struck and the frequency at which it vibrates and produces the loudest sound is recorded.

## 3. Why is the resonant frequency of a tuning fork important?

The resonant frequency of a tuning fork is important because it is used as a standard for tuning musical instruments. It is also used in scientific experiments to study the behavior of waves and vibrations.

## 4. Can the resonant frequency of a tuning fork be changed?

Yes, the resonant frequency of a tuning fork can be changed by altering its physical characteristics, such as its length or material. Additionally, the resonant frequency can be affected by external factors such as temperature and humidity.

## 5. What are some real-world applications of the resonant frequency of a tuning fork?

The resonant frequency of a tuning fork has many real-world applications, including in music and sound therapy. It is also used in scientific research for precise measurements and in engineering for testing the strength and durability of materials.

• Introductory Physics Homework Help
Replies
15
Views
7K
• Introductory Physics Homework Help
Replies
6
Views
4K
• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
33
Views
3K
• Introductory Physics Homework Help
Replies
22
Views
4K
• Introductory Physics Homework Help
Replies
12
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
4K