What is the Minimum Depth of Water for Standing Sound Waves in a Drinking Glass?

AI Thread Summary
The discussion focuses on determining the minimum depth of water in a drinking glass that enhances sound waves emitted by a speaker at a frequency of 3.75 kHz. The calculations involve using the speed of sound in air (343 m/s) and the relationship between wavelength and frequency to find the depth where standing sound waves form. Initial attempts to calculate the depth yielded incorrect results, prompting suggestions to use higher odd values of n for the wavelength formula. Participants also emphasized the importance of using the correct speed of sound and maintaining significant figures to avoid rounding errors. The problem illustrates the sensitivity of sound wave behavior to physical parameters in a confined space.
4Phreal
Messages
17
Reaction score
0

Homework Statement



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



Homework Equations



fn = nv/4L
v=fλ


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:
Physics news on Phys.org
4Phreal said:

Homework Statement



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



Homework Equations



fn = nv/4L
v=fλ


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

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

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.
 
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! :smile:]
 
Last edited:
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top