# Wavelength in a closed column

1. Jan 29, 2014

### 4Phreal

1. The problem statement, all variables and given/known data

We place a speaker near the top of a drinking glass. The speaker emits sound waves with a frequency of 3.75 kHz. The glass is 14.1 cm deep. As I pour water into the glass, I find that at certain levels the sound is enhanced due to the excitation of standing sound waves in the air inside the glass. Find the minimum depth of water at which this occurs (distance from surface of water to bottom of glass). The standing sound wave has a node at the surface of the water and an antinode at the top of the glass. Assume that the antinode is exactly at the top of the glass. The speed of sound in air is 343 m/s

2. Relevant equations

fn = nv/4L
v=fλ

3. The attempt at a solution

The distance between a node and an antinode is wavelength/4, so I calculate wavelength via lambda=v/f, converting 3.75 kHz to 3750 Hz. This gives wavelength equal to 0.091466667m. I then take 14.1 cm, convert it to 0.141 m, and subtract the 0.091466667/4 from it to get the depth of the water, which is 0.118133333 m, or 11.8 cm. Which is wrong.
I next try 5λ/4 (using n=5) and get 0.027 m, or 2.7 cm which is also wrong, but 7λ/4 yields a negative answer, and 3λ/4 is gives an answer larger than 5λ/4

Last edited: Jan 29, 2014
2. Jan 30, 2014

### collinsmark

That would give you maximum depth of water rather than the minimum.

Sounds to me like you have the right idea. Try to make n as large as possible (while keeping it odd) without going over the 14.1 cm depth of the glass.

Are you sure you are using the correct speed of sound that your coursework requires? I see that you are using 343 m/s. But the speed of sound can vary considerably depending on altitude mostly, but also a little on other factors such as barometric pressure, humidity, etc. Make sure you are using the number you are supposed to use, whatever that may be.

Other than that, my only advice is to make sure you keep plenty of significant figures in your calculation up to the end to avoid compounded rounding errors.

3. Jan 30, 2014

### collinsmark

Also, for what it's worth, this problem is very sensitive to to the wavelength, and thus the speed of sound used.

An small percentage error in the wavelength (or speed of sound) will translate to an error of a much larger percentage in the final answer.

[Edit: Oh, and welcome to Physics Forums! ]

Last edited: Jan 30, 2014