# Resonance box with tuning fork, standing wave

late347

## Homework Statement

There is a resonance box with one end cloesd and the other end open. The box reinforces the sound of the tuning fork. That sound has frequency of 440 Hz
sound velocity is 340 m/s

a.) What is the basis of the phenomenon in question?
b.) define the shortest possible length of the resonance box

v= λ * f

## The Attempt at a Solution

I winced a little bit when I saw that a.) portion appear in my exam. I decide to skip this question on the basis of not being sure what kind of explanation was "enough"

We had an elective type of exam, choose 5 out of 6 problems. This one I skipped. I feel that I did have a rudimentary understanding of the situation and the basis of the phenomenon, but I was unsure if my understanding was
1) without error
2) good enough of an explanation

But the second question about the length is easier to answer in a more straightforward and mathematical manner. So, I will probably go with reverse order, first B and then A.

b.)

Find the wavelength first

## \lambda = \frac {v} {f} ##
## \lambda = 0.7727m ##

We know that between a node and another node there is ##\frac {\lambda} {2}##
In the resonance box, one wall of the box is closed with a node point, and the other side is open with an antinode.

## \frac {\lambda} {4} = l ##
## l = \frac {0.7727m} {4} = 0.2m ##

a.) explain the basis of the phenomenon

sound is a longitudinal wave, which requires a medium to travel in. The basis of the phenomenon appears to be such that a standing wave is created in the resonance box. And the node point will be the closed-off wall of the box, and the antinode is the open end.
I had a hazy recollection that the antinode is actually defined by the currently prevalent meteorological conditions, since the medium in this case appears to be air.

Standing waves themselves seem to be caused by a superposition of two waves which traveled in opposite directions. If the two waves have same amplitude, wavelength and frequency then it would seem that the standing wave is created.

Precisely how this happens in a resonance box, is a little bit unclear to me. It would seem that the tuning fork is the culprit for creating a sound wave, which is reinforced by the resonance box.

Homework Helper
I'd have given you full marks Homework Helper
Gold Member
Everything sounds clear to me in your explanation. Do you understand why they asked for the shortest possible distance? There are other distances which would also resonate at the frequency (meaning there will be an antinode at the open end, and node at the closed end). How do these distances compare to the shortest distance?

lychette
The resonance length is fractionally longer than the physical length of the tube.
An 'end correction' must be added to the physical length of the tube, this can be determined experimentally.

Maria B
Hey all! I am experimenting for a sound art project with tuning forks and resonance boxes. I have a complementary question to this thread:

If I have a resonance box specific for 1/4 λ of a given frequency of a tuning fork, can I add several tuning forks (all the same size and freq)?