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Chapter 16: Problem 30

Singing in the Shower. A pipe closed at both ends can have standing waves inside of it, but you normally don't hear them because little of the sound can get out. But you \(can\) hear them if you are \(inside\) the pipe, such as someone singing in the shower. (a) Show that the wavelengths of standing waves in a pipe of length \(L\) that is closed at both ends are \(\lambda$$_n =\) \(2L /n\) and the frequencies are given by \({f_n}\) \(= n$$v/2L = nf$$_1\), where \(n =\) 1, 2, 3, . . . . (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 m tall. Are these frequencies audible?

Short Answer

Expert verified
The fundamental frequency is 68.6 Hz and overtones are 137.2 Hz and 205.8 Hz, all audible.

Step by step solution

01

Understanding Standing Waves in a Pipe Closed at Both Ends

A pipe closed at both ends has nodes at both ends. The wavelength of the standing wave in such a pipe can be derived from the condition that the pipe length is an integer multiple of half wavelengths: \[ L = \frac{n\lambda}{2} \]Solving for wavelength \( \lambda \), we get: \[ \lambda = \frac{2L}{n} \] where \( n = 1, 2, 3, \ldots \)
02

Calculating Frequency of Standing Waves

The frequency \( f \) of a wave is related to its wavelength \( \lambda \) and wave speed \( v \) by: \[ f = \frac{v}{\lambda} \]Substituting \( \lambda = \frac{2L}{n} \) from the previous step, the frequency \( f_n \) becomes: \[ f_n = \frac{nv}{2L} = nf_1 \] where \( f_1 \) is the fundamental frequency with \( n = 1 \).
03

Calculate the Fundamental Frequency

The fundamental frequency \( f_1 \) is calculated using:\[ f_1 = \frac{v}{2L} \]Given: length of the shower \( L = 2.50 \) m and speed of sound \( v \approx 343 \) m/s, so\[ f_1 = \frac{343}{2 \times 2.50} \approx 68.6 \text{ Hz} \]
04

Find Frequencies of the First Two Overtones

Overtones are the higher harmonic frequencies. The first overtone will have \( n = 2 \):\[ f_2 = 2 \times 68.6 \approx 137.2 \text{ Hz} \]The second overtone will have \( n = 3 \):\[ f_3 = 3 \times 68.6 \approx 205.8 \text{ Hz} \]
05

Assessing Audibility of the Frequencies

Human hearing typically ranges from about 20 Hz to 20,000 Hz. The calculated frequencies are approximately 68.6 Hz (fundamental), 137.2 Hz (first overtone), and 205.8 Hz (second overtone). All of these frequencies fall within the audible range.

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Key Concepts

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

Pipe Acoustics
When we talk about pipe acoustics, we are diving into the science of how sound behaves in cylindrical structures, like a pipe. Pipes can amplify sound by acting as resonant cavities, which means they reflect sound waves and create standing waves. This is particularly fascinating because pipes can be closed at both ends, like in our example of a shower, creating a specific type of resonance. Standing waves within such a pipe have nodes, or points of no movement, at both ends.

This situation arises because the pipe length is an integer multiple of half wavelengths of the sound 鈥� essentially dividing the pipe into segments where waves reinforce each other. By understanding this, you can predict how sound waves will behave in any enclosed space, explaining why a bathroom can be a good place to sing, enhancing our existing vocal capabilities by naturally increasing sound resonance.
  • Pipes closed at both ends act as resonant chambers.
  • Standing waves have nodes at both ends of the pipe.
  • The length of the pipe is a critical factor in determining the wavelengths.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which a system like a pipe supports standing waves. For a pipe closed at both ends, this concept is key to understanding what notes or pitches the pipe can naturally enhance. The fundamental frequency, often signified as \( f_1 \), is determined by the length of the pipe and the speed of sound in air.

Mathematically, the fundamental frequency for a pipe can be calculated with the formula:\[ f_1 = \frac{v}{2L} \]where \( v \) is the speed of sound (assumed to be approximately 343 m/s under typical conditions) and \( L \) is the length of the pipe.
  • The fundamental frequency is the lowest natural frequency of vibration.
  • For a 2.50 m tall shower, the fundamental frequency is about 68.6 Hz.
  • This frequency is crucial since it forms the basis for calculating higher harmonics.
Audible Frequencies
Audible frequencies refer to the range of sound frequencies that the average human ear can detect. Generally, this range falls between 20 Hz and 20,000 Hz. Understanding whether a frequency is audible or not helps to determine if it can be heard during real-life scenarios, such as singing in the shower.

In our example, we calculated the fundamental frequency to be approximately 68.6 Hz and the first two overtones to be about 137.2 Hz and 205.8 Hz. All these frequencies comfortably sit within the audible range for most people. This means you can definitely hear the notes that are reinforced by the natural resonance of the pipe or shower. It's not just about hearing but experiencing how certain environments naturally augment different sound frequencies, providing a richer auditory experience.
  • Human hearing ranges from about 20 Hz to 20,000 Hz.
  • The calculated frequencies are all audible.
  • This enhances the acoustic experience within the pipe or shower.

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Most popular questions from this chapter

You are trying to overhear a juicy conversation, but from your distance of 15.0 m, it sounds like only an average whisper of 20.0 dB. How close should you move to the chatterboxes for the sound level to be 60.0 dB?

A turntable 1.50 m in diameter rotates at 75 rpm. Two speakers, each giving off sound of wavelength 31.3 cm, are attached to the rim of the table at opposite ends of a diameter. A listener stands in front of the turntable. (a) What is the greatest beat frequency the listener will receive from this system? (b) Will the listener be able to distinguish individual beats?

While sitting in your car by the side of a country road, you are approached by your friend, who happens to be in an identical car. You blow your car's horn, which has a frequency of 260 Hz. Your friend blows his car's horn, which is identical to yours, and you hear a beat frequency of 6.0 Hz. How fast is your friend approaching you?

A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated tone of exactly that frequency and hears a beat frequency of 3 Hz, which increases to 4 Hz when she tightens her violin string slightly. (a) What was the frequency of the note played by her violin when she heard the 3-Hz beats? (b) To get her violin perfectly tuned to concert A, should she tighten or loosen her string from what it was when she heard the 3-Hz beats?

The siren of a fire engine that is driving northward at 30.0 m/s emits a sound of frequency 2000 Hz. A truck in front of this fire engine is moving northward at 20.0 m/s. (a) What is the frequency of the siren's sound that the fire engine's driver hears reflected from the back of the truck? (b) What wavelength would this driver measure for these reflected sound waves?
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