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Chapter 16: Problem 63

(a) Determine the first three normal-mode frequencies for a pipe of length \(L\) that is closed at both ends. Explain your reasoning. (b) Use the results of part (a) to estimate the normal-mode frequencies of a shower stall. Explain the connectiva between these frequencies and the observation that your singing voice probably sounds better in the shower, especially when you sing at certain frequencies.

Short Answer

Expert verified
The first three normal-mode frequencies for a closed pipe are \( f_1 = \frac{v}{2L} \), \( f_2 = \frac{v}{L} \), \( f_3 = \frac{3v}{2L} \).

Step by step solution

01

Understanding the Problem

To begin, we need to determine the normal-mode frequencies for a pipe closed at both ends. A pipe closed at both ends supports standing waves, with nodes at each end. We can express these standing wave frequencies using harmonic formulas.
02

Mathematical Representation of the Frequencies

For a pipe of length \(L\) that is closed at both ends, the normal-mode frequencies can be represented using the formula: \[ f_n = \frac{nv}{2L} \] where \(n\) is the harmonic number (integer), \(v\) is the speed of sound, and \(L\) is the length of the pipe. With \(n\) taking values 1, 2, 3,...
03

Calculating the First Three Frequencies

We calculate the first three normal-mode frequencies by using \(n = 1, 2,\) and \(3\): 1. First frequency (\(n=1\)): \( f_1 = \frac{v}{2L} \)2. Second frequency (\(n=2\)): \( f_2 = \frac{v}{L} \)3. Third frequency (\(n=3\)): \( f_3 = \frac{3v}{2L} \)
04

Applying to a Shower Stall

A shower stall can be approximated as a pipe closed at both ends, so we can use the same frequency formula. Estimate \(L\) as the height or depth, and calculate \(f_1, f_2,\) and \(f_3\) using the speed of sound \(v\approx343\,\text{m/s}\).
05

Explaining Sound Quality

These calculated frequencies correspond to the natural resonant frequencies of the shower stall. Singing at these frequencies causes resonance, which enhances sound quality and volume, making your voice sound better.

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Key Concepts

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

Standing Waves
Standing waves are a fascinating phenomenon where waves stay in a constant position. Imagine a wave that doesn鈥檛 seem to move left or right. This happens when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other.
They form specific patterns called nodes and antinodes:
  • Nodes are the points where there鈥檚 no movement. In a standing wave, they appear as still points.
  • Antinodes, on the other hand, are the spots where the movement is maximum. These are the oscillating parts of the wave.
Standing waves occur in many everyday objects, like musical instruments. They are key to understanding how sound is produced and manipulated for different effects.
Pipe Closed at Both Ends
A pipe closed at both ends is an interesting structure for waves because it supports standing waves in a very particular way. The ends of the pipe act as nodes, where the wave must have zero displacement.
This restricts the possible wave patterns that can form within the pipe. As a result, only specific frequencies can exist. These frequencies correspond to standing wave patterns inside the pipe.
  • The length of the pipe plays a crucial role in determining these frequencies.
  • Moreover, the speed of sound in the pipe will affect how quickly these frequencies can form.
Understanding pipes closed at both ends helps in musical acoustics and in predicting how natural structures, like a shower stall, enhance vocal sounds.
Resonant Frequencies
Resonant frequencies are those at which a system naturally oscillates with greater amplitude. In simpler terms, they are the frequencies at which an object naturally wants to vibrate.
For a pipe or a space closed at both ends, these frequencies map directly to the standing wave frequencies.
  • When a sound wave at a resonant frequency is introduced, it aligns perfectly with the natural frequencies of the system.
  • This can lead to a significant increase in the amplitude of the wave, producing a clearer and louder sound.
In everyday life, resonant frequencies explain why certain sounds are amplified in places with closed ends, like a pipe or a shower stall. The latter is why you might feel your singing improves when inside!
Harmonic Frequencies
Harmonic frequencies refer to frequencies obtained by multiplying a fundamental frequency by integers. In a pipe closed at both ends, they are simply multiples of the fundamental frequency.
Let's break it down:
  • The fundamental frequency (first harmonic) is the lowest frequency at which a system can oscillate.
  • The second harmonic is twice the fundamental frequency, the third harmonic is three times the fundamental frequency, and so on.
Calculating harmonic frequencies in a closed pipe helps predict the sequence of notes or sounds that the pipe can produce. Each harmonic adds a layer of depth and complexity to the sounds we hear, much like the enriching effect harmonics have in music and speech, especially observed in closed spaces like shower stalls.

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Most popular questions from this chapter

Horseshoe bats (genus Rhinolophus) emit sounds from their nostrils and then listen to the frequency of the sound reflected from their prey to determine the prey's speed. (The "horseshoe" that gives the hat its name is a depression around the nostrils that acts like a focusing mirror, so that the hat emits sound in a narrow beam like a flashlight.) A Rhinolophus flying at speed \(v_{\text { tot }}\) emits sound of fre-quency \(f_{\text { but }}\) ; the sound it hears reflected from an insect flying toward it has a higher frequency \(f_{\text { rent }}(\text { a) Show that the speed of the insect is }\) where \(v\) is the speed of sound. (b) If \(f_{\mathrm{bat}}=80.7 \mathrm{kHz}, \quad f_{\mathrm{rell}}=\) \(83.5 \mathrm{kHz},\) and \(v_{\mathrm{bat}}=3.9 \mathrm{m} / \mathrm{s},\) calculate the speed of the insect.

Two identical taut strings under the same tension \(F\) produce a note of the same fundamental frequency \(f_{0}\) . The tension in one of them is now increased by a very small amount \(\Delta F\) . (a) If they are played together in their fundamental, show that the frequency of the beat produced is \(f_{\text { keet }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 \(\mathrm{Hz}\) . One of the strings is retuned by increasing its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?

(a) In a liquid with density 1300 \(\mathrm{kg} / \mathrm{m}^{3}\) , longitudinal waves with frequency 400 \(\mathrm{Hz}\) are found to have wavelength 8.00 \(\mathrm{m}\) . Calculate the bulk modulus of the liquid. (b) A metal bar with a length of 1.50 \(\mathrm{m}\) has density 6400 \(\mathrm{kg} / \mathrm{m}^{3}\) . Longitudinal sound waves take \(3.90 \times 10^{-4} \mathrm{s}\) to travel from one end of the bar to the other. What is Young's modulus for this metal?

Two identical loudspcakers are located at points \(A\) and \(B\) , 2.00 \(\mathrm{m}\) apart. The loudspeakers are driven by the same amplifier and produce sound waves with a frequency of 784 \(\mathrm{Hz}\) . Take the speed of sound in air to be 344 \(\mathrm{m} / \mathrm{s}\) . A small microphone is moved out from point \(B\) along a line perpendicular to the line connecting \(A\) and \(B(\text { line } B C \text { in Fig. } 16.44)\) (a) At what distances from \(B\) will there be destructive interference? (b) At what distances from \(B\) will there be constructive interference? (c) If the frequency is made low enough, there will be no positions along the line \(B C\) at which destructive interference occurs. How low must the frequency be for this to be the case?

Two loudspeakers, \(A\) and \(B,\) are driven by the same amplifier and emit sinusoidal waves in phase. Speaker \(B\) is 12.0 \(\mathrm{m}\) to the right of speaker \(A\) . The frequency of the waves emitted by each speaker is 688 \(\mathrm{Hz}\) You are standing between the speakers, along the line connecting them, and are at a point of constructive interference. How far must you walk toward speaker \(B\) to move to a point of destructive interference? effects like those in parts (a) and (b) are almost never a factor in listening to home stereo equipment. Why not?
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