Sorry for the repost, i made a typo in the original post's subject.
A dry water well is 1.5 m in radius and 5.0 m deep.
In order from lowest to highest, the fifth resonant frequency of this well is closest to ...
I know that it is an open-closed system, but how would we take into account the amplitude, which is the radius given for this question?
To determine the frequency of resonance in a closed-open system, you can use the formula:
f = v/2L,
where f is the frequency of resonance, v is the speed of sound, and L is the length of the air column. In this case, the air column is equivalent to the depth of the well.
First, you need to determine the length of the air column. Since the well is dry, we can assume it is filled with air. Therefore, the length of the air column in the well is equal to its depth, which is given as 5.0 m.
Next, you need to know the speed of sound in air. The speed of sound depends on the temperature, so you may need to refer to a table or use an approximation. At room temperature (around 20 °C), the speed of sound in air is approximately 343 m/s.
Now, you can plug in the values into the formula and solve for the resonance frequency:
f = 343/(2 * 5.0).
Calculating this expression will give you the fifth resonant frequency of the well. However, we need to consider the amplitude, which is represented by the radius.
In this case, the amplitude, represented by the radius of 1.5 m, does not directly affect the resonance frequency in this closed-open system. The radius of the well only affects the intensity or loudness of the sound produced, not its frequency.
Therefore, you can use the formula mentioned earlier without considering the amplitude when calculating the fifth resonant frequency of the well.