A 680 Hz tuning fork is held over a long vertical tube full of water. If the tube is 1.00 m long, how many resonances will be heard as the tube is allowed to empty from the bottom?
resonances will be heard each odd number of quarter wavelengths.
Determine the wavelength (f*lambda=wavespeed), divide it by four, then figure out the odd multiples that are less than 1 meter.
12
To determine the number of resonances that will be heard as the tube is being emptied, we need to consider the relationship between the frequency of the tuning fork and the length of the tube.
The fundamental frequency (or first harmonic) of a closed tube (such as the one described in the question) can be calculated using the formula:
f = v / (2L)
Where:
f is the frequency of the sound wave (in Hz),
v is the speed of sound in air (approximately 343 m/s at room temperature), and
L is the length of the tube (in meters).
In this case, the frequency of the tuning fork is given as 680 Hz, and the length of the tube is 1.00 m.
Let's substitute these values into the equation:
680 Hz = 343 m/s / (2 × 1.00 m)
Now, we can solve for the speed of sound in air:
680 Hz × 2 × 1.00 m = 343 m/s
1360 m = 343 m/s
Simplifying further:
1360 / 343 ≈ 3.97
This means that the closest number of resonances, or the number of times the tube will resonate, will be around 3.97. However, since we cannot have a fraction of a resonance, we need to round this value to the nearest whole number.
Hence, as the tube is being emptied, approximately 4 resonances will be heard.