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Chapter 5: Problem 18

Measuring the relative phase at the two ends of an open tube. Suppose someone has taken a long hoselike tube, coiled it up in a box, and let one open end stick out one side of the box and the other out the other. You are not allowed to see how much of the tube is coiled inside the box. By adding a small tuning trombone to a protruding end, you find that you get a resonance at \(523.3\) cps from your tuning fork. That means that the total length is either \(\frac{1}{2} \lambda\), or \(\lambda\), or \(\frac{A}{2} \lambda\), or \(\ldots . .\) How can you find out whether the tube is an odd or even number of half-wavelengths? Hold two vibrating forks at one end of the tube and listen to the beats. Get the rhythm in your head so that if you remove one fork momentarily and then replace it (without disturbing the continued vibrations of both forks), you can tell that the beat maximum comes "on the beat" (in musical jargon) just where it should be. Practice several times so that you can skip a beat, count beats in your head, and come back in step when you replace the fork. (You can adjust the rubber-band loading to get a convenient beat frequency. If you find all this difficult, you can use a metronome.) Now! This time, instead of replacing the (momentarily) removed fork at the same end of the tube, carry it to the other end. Again listen for the beats. (Both forks have continued vibrating all this time.) Do they come back "on the beat," or do they come back "on the off-beat"? Depending on the experimental result, you should be able to decide whether the tube is an odd or even number of half-wavelengths. Predict the answer; then try the experiment with your half-wavelength tube. (Make another tube one wavelength long to get the opposite result.)

Short Answer

Expert verified
Listen for beats at both tube ends: on-beat suggest even half-wavelengths, off-beat suggest odd half-wavelengths.

Step by step solution

01

Understanding Resonance

We know that for resonance to occur, the length of the tube must be a multiple of half-wavelengths. Given the frequency of 523.3 cps (cycles per second), this suggests that the length of the tube could be \(\frac{1}{2}\lambda\), \(\lambda\), \(\frac{3}{2}\lambda\), etc. Determine possible lengths based on these calculations.
02

Preparing for Beat Measurement

With two tuning forks at one end, measure the beat frequency when both are vibrating. Ensure you can count and replace one fork while maintaining synchronization. This serves as a control to determine if the tube length is an odd or even multiple.
03

Conducting Beat Observation

Relocate one fork to the other end of the tube and listen for beats. If the beats are synchronized, the phase between the two ends is the same, indicating an even number of half-wavelengths. If they are off-beat, it's an odd number, showing a phase difference of \(\pi\) radians.
04

Analyzing Results

Based on whether the beats align or are off, decide if the tube is an odd or even number of half-wavelengths. Aligning beats suggest even multiples, while off-beat indicates odd multiples.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Resonance
Resonance is a fascinating physical phenomenon that occurs when an object vibrates at its natural frequency due to an external force. In the scenario of a tube and tuning fork, resonance happens when the length of the tube matches a certain multiple of the wavelength of the sound wave produced by the tuning fork. For example, if the wavelength of the sound is such that it perfectly fits into the tube's length (or a multiple of it), the sound waves will reinforce each other, producing a stronger, resonating sound. This amplification means the sound is "in tune" with the tube's natural frequency, resulting in a clearer, louder tone apart from ordinary sound. Conditions for resonance can be calculated by considering the wave relations:
  • If the length of the tube is \(\frac{1}{2}\lambda\), only a quarter of the wave fits in one side, making it fundamental.
  • If it matches \(\lambda\), or half of the full wave, it's a single complete set of that tone.
  • Higher harmonics like \(\frac{3}{2}\lambda\), \(2\lambda\), etc., create more complex and richer sound waves due to layers of vibrations.
Tuning Forks
Tuning forks are precision instruments used to produce a specific frequency when struck. Each tuning fork is designed to generate sound waves at a precise frequency, and they are often used to tune musical instruments or conduct experiments related to sound waves, like in the tube experiment. In the given exercise, tuning forks serve a critical role by providing a consistent sound wave of known frequency (523.3 cps). This allows for an investigation into the tube's resonant frequency and length by observing the resulting wave interactions. When using two tuning forks, interesting phenomena occur, notably the creation of beats, a rhythmic pulsing resulting from slightly different frequencies. Adjusting one or both forks finely changes the interaction of frequencies, offering insight into the exact lengths involved, and further aiding in deducing the tube's characteristics.
Phase Difference
Phase difference refers to the difference in phase between two points on a wave or between two waves of the same frequency. In this context, the phase difference between waves at two ends of the tube can help determine if the tube's length is an odd or even number of half-wavelengths. Understanding phase difference is crucial when exploring wave mechanics, as it forms the basis for concepts like constructive and destructive interference.
  • When waves are in phase (same phase shift), they reinforce each other, leading to a stronger combined wave.
  • A phase difference of \(\pi\) radians (half a wavelength) would mean one wave crest aligns with the other's trough, potentially canceling each other out.
This destructive interference results in beats, a practical outcome richly observed in the tube and tuning forks exercise. This phase difference leads to the key experimental result on whether the tube is an odd or even multiple of half-wavelengths.
Beat Frequency
Beat frequency is the number of beats per second, resulting from the interference of two sound waves of slightly different frequencies. When two frequencies are different but close, they create a resultant sound that fluctuates in loudness, known as beats.In acoustic terms, these fluctuations are the rate at which the two frequencies diverge or converge over time. The beat frequency can be calculated using:\[ f_{beat} = |f_1 - f_2| \]where \( f_1 \) and \( f_2 \) are the frequencies of the two sound waves. Applying this to the tuning forks in the tube experiment, observing beat frequency helps identify phase differences and whether length equates to odd or even multiples of the wave's half-length. Beats indicate either canceling phases (odd multiples, off-beat) or harmonizing phases (even multiples, on-beat), thus acting as auditory clues to underlying physical wave and tube properties.

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Most popular questions from this chapter

Nonreflecting coating. A glass lens has been coated with a nonreflecting coating that is one quarter-wavelength in thickness tn the coating for light of cactum wavelength \(\lambda_{0}\). The index of refraction of the coating is \(\sqrt{n}\); that of the glass is \(n\). Take the index of refraction to be constant, independent of frequency, over the visible frequency spectrum. Let \(I_{\text {ref }}\) denote the time-averaged reflected intensity and \(I_{0}\) the incident intensity, for light at normal incidence. Show that the fractional reflected intensity has the following dependence on the wavelength of the incident light: $$ \frac{I_{\text {rot }}}{I_{0}}=4\left[\frac{1-\sqrt{n}}{1+\sqrt{n}}\right]^{2} \sin ^{2} \frac{1}{2} \pi\left(\frac{\lambda_{0}}{\lambda}-1\right) $$ where \(\lambda\) is the vacuum wavelength of the incident light. Take \(n=1.5\) for glass. Suppose \(\lambda_{0}=5500 \mathrm{~A}\) (green light), Then \(I_{\mathrm{ret}}\) is zero for green. What is \(I_{\mathrm{ret}} / I_{0}\) for blue light of vacuum wavelength \(4500 \mathrm{~A}\) ? What is it for red light of vacuum wavelength \(6500 \AA ?\)

Termination of waves on a string. (a) Suppose you have a massless dashpot having two moving parts 1 and 2 that can move relative to one another along the \(x\) direction, which is transverse to the string direction \(z\). Friction is provided by a fluid that retards the relative motion of the two moving parts. The friction is such that the force needed to maintain relative velocity \(\dot{x}_{1}-\dot{x}_{2}\) between the two moving parts is \(Z_{d}\left(\dot{x}_{1}-\dot{x}_{2}\right)\), where \(Z_{i} d\) is the impedance of the dashpot. The input (part 1 ) is connected to the end of a string of impedance \(Z_{1}\) stretching from \(z=-\infty\) to \(z=0 .\) The output (part 2 ) is connected to a string of impedance \(Z_{2}\) that extends to \(z=+\infty\). Show that a wave incident from the left experiences an impedance at \(z=0\) which is the same as that it would experience if connected to a "load" consisting of a string stretching from \(z=0\) to \(+\infty\) and having impedance \(Z_{L}\) given by $$ \mathrm{Z}_{L}=\frac{\mathrm{Z}_{\mathrm{d}} \mathrm{Z}_{2}}{\mathrm{Z}_{d}+\mathrm{Z}_{2}}, \quad \text { that is, } \quad \frac{1}{\mathrm{Z}_{L}}=\frac{1}{\mathrm{Z}_{d}}+\frac{1}{\mathrm{Z}_{2}} \text { . } $$ Thus it is as if the dashpot and string 2 were impedances connected "in parallel" and driven by the incident wave. (b) Show that if string \(Z_{2}\) extends only to \(z=\frac{1}{4} \lambda_{2}\), where \(\lambda_{2}\) is the wavelength in medium 2 (assuming we have a harmonic wave with a single frequency), and there is terminated by a dashpot of zero impedance (frictionless), the wave incident at \(z=0\) is perfectly terminated. Show that the output connection of the dashpot at \(z=0\) cannot tell whether it is connected to a string of infinite impedance or is instead connected to a quarter-wavelength string that is "short-circuited" by a frictionless dashpot at \(z=\frac{1}{4} \lambda_{2 \cdot}\) In either ease the output connection remains at rest.

Transitory standing waves on a slinky. Attach one end of a stinky to a telephone pole or something. Hold the other end. Stretch the slinky out to \(30 \mathrm{ft}\) or so. Shake the end of the slinky about 3 or 4 times as rapidly as you can. \(\mathrm{A}^{\text {" } \text { wave packet" }}\) " is thus propagated down the slinky. After you have sufficiently enjoyed following packets back and forth, try something new: This time, keep your attention fixed on a region near the fixed end of the slinky. As the packet comes in, reflects, and returns, you should see transitory standing waves during the time interval in which the incident and reflected wave packets overlap. (It may help to fix both ends of the slinky so that you can watch the process at close range at your end of the slinky.) That should help to convince you that a standing wave can always be regarded as the superposition of two traveling waves traveling in opposite directions.

Impedance matching by "tapered" index of refraction. Suppose you want to match optical impedances between a region of index \(n_{1}\) and a region of index \(n_{2}\), and you want to expend a total distance \(L\) in the impedance-matching transition region. What is the optimum \(z\) dependence of the index \(n\) between the two regions? Is it exponential? Why not?

Reflection from glass. A flat slab of glass reflects about 8\% of incident light intensity for normal incidence, \(4 \%\) in intensity being reflected from each surface. An ordinary silvered mirror reflects more than \(90 \%\) of the visible light. Take a mirror and a single clean piece of glass (a microscope slide, for example). Compare the reflection from the mirror and from the slide with the two held close together so you can see both reflections at once. Look at the reflection of a broad light source like an incandescent bulb, or a piece of white paper, or a patch of sky. Compare the reAlectivity of the slide and that of the mirror at near-normal incidence. Now do the same at near- grazing incidence. At near-grazing incidence, the source, mirror reflection, and slide reflection should be nearly indistinguishable; i.e., you get nearly \(100 \%\) reflection at near-grazing ineidence. At near-normal incidence, the glass should be noticeably dimmer than the mirror. Next take four clean microscope slides. Lay them on top of one another in a series of "steps," with the first slide giving a "floor, " the second slide giving the first "step," and the remaining two slides giving a second "step" of double height. Thus, you can compare at the same time reflection at near- normal incidence from one slide, two slides, and four slides. Look at a broad source (the sky) reflected at near-normal incidence. Neglecting the complications of internal reflections, you should transmit about \(0.92\) through each slide. Thus, four slides should transmit \((0.92)^{4}=0.72\) and reflect \(1-(0.92)^{4} \approx 0.28\) Now make a pile of about a dozen clean slides; they should reflect \(1-(0.92)^{12}=0.64\). Compare with the mirror. Suppose the formula keeps working (and the slides are elean). How many slides will equal one good mirror if the mirror reflects \(93 \%\) of the intensity? Try it-compare the stack of slides with the mirror at near-normal incidence. Also look directly at the source through the stack to see the transmitted light. (It takes about 32 slides, according to the formula. Needless to say, they should be free of fingerprints.) (Microscope slides cost about 35 cents per dozen. Three dozen slides is a good investment for home experiments.)
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