A 1.0- m-tall vertical tube is filled with 20^∘ C water. A...

A $1.0-$ m-tall vertical tube is filled with $20^{\circ} \mathrm{C}$ water. A tuning fork vibrating at $580 \mathrm{Hz}$ is held just over the top of the tube as the water is slowly drained from the bottom. At what water heights, measured from the bottom of the tube, will there be a standing wave in the tube above the water?

Key Concepts

Standing Waves

Standing waves are the result of the interference between two waves of the same frequency traveling in opposite directions, producing a waveform that appears to be stationary. In acoustics, these waves form in air columns when sound reflects repeatedly off the boundaries, creating fixed patterns of nodes (points of no displacement) and antinodes (points of maximum displacement).

Acoustic Resonance

Acoustic resonance occurs when an air column or cavity is driven at one of its natural frequencies, leading to a large amplitude oscillation. This amplification happens because the frequency of the driving source matches one of the resonant frequencies of the system, efficiently transferring energy into the system’s standing wave pattern.

Boundary Conditions in Air Columns

In air columns, boundary conditions determine where nodes and antinodes form. For a tube with one end closed (or effectively closed, such as by a water surface) and the other open, the closed end enforces a displacement node while the open end supports a displacement antinode. These conditions are critical for determining the possible standing wave patterns and the resonant frequencies of the tube.

Resonant Frequencies and Harmonics in Closed Tubes

For tubes that are closed at one end and open at the other, resonance occurs at specific frequencies corresponding to odd multiples of the quarter-wavelength. The effective length L of the resonator satisfies L = (2n – 1)?/4, where n is an integer (1, 2, 3, ...). This relationship implies that only the odd harmonics are present, which is a distinctive feature of closed tube resonators.

Relationship between Frequency, Wavelength, and Speed

The fundamental relationship between frequency (f), wavelength (?), and the speed of sound (v) is given by the equation ? = v/f. This equation is essential in acoustics as it links the physical dimensions of the resonator with the frequency of the sound wave, allowing for the calculation of resonant conditions when one of these quantities changes.

Transcript

00:01 I've drawn a diagram of this situation here.

00:05 I don't know why it went down ever so slightly.

00:12 Regardless, this is an open -closed situation because the top is going to be open and the bottom is going to be contacting the water.

00:28 In an open -closed situation, f sub -m equals m -m -v -m -v.

00:45 Over 4l.

01:01 But in this case, l is going to be replaced by the length of air in the pipe, which is going to be l minus h, where m equals 1, 3, 5, 7.

01:28 So m needs to be an odd number.

01:32 So if we take our frequency, which is f, and multiply by 4, divide by v, that's going to give me m over l minus h.

02:06 All right.

02:07 The point is that m has to be an integer.

02:12 So i actually want to solve this for h, and then i just want to substitute in values for m.

02:25 So let's keep going.

02:28 For f, so i want to go maybe a little bit slightly different direction here.

02:41 For f, l minus h, equals mv.

02:52 So l minus h is going to be mv over for f.

03:01 And so so h is going to be l minus mv over 4f.

03:11 So if we let m equal 1, and i put that in a calculator, l is 1 minus m is 1, v 343 over 4f, 580 is f.

03:43 So i get, first of all, when m is 1, is 0 .85 meters...

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