- #1

KYPOWERLIFTER

- 19

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## 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.