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Chapter 16: Problem 39

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

Short Answer

Expert verified
(a) 437 Hz; (b) Tighten the string.

Step by step solution

01

Understanding Beats and Frequency

When two sounds are played together and have slightly different frequencies, the interference between the two produces a phenomenon known as 'beats'. The beat frequency is equal to the absolute difference between the two frequencies. Mathematically, this is described as \( f_{beat} = |f_1 - f_2| \). In our exercise, the electronically generated tone is at 440 Hz and the beat frequency initially heard is 3 Hz.
02

Calculating the Initial Frequency of the Violin

Given the beat frequency of 3 Hz and knowing that one tone is at 440 Hz, use the beat frequency formula to find the violin's frequency. There are two possible cases: 1) \( f_{violin} = 440 \, \text{Hz} + 3 \, \text{Hz} = 443 \, \text{Hz} \), or 2) \( f_{violin} = 440 \, \text{Hz} - 3 \, \text{Hz} = 437 \, \text{Hz} \). Both values need to be considered to determine the correct frequency.
03

Impact of Tightening the String

When the violinist tightens the string, the frequency increases. After tightening slightly, the beat frequency changes to 4 Hz. A higher beat frequency means the frequency difference has increased, confirming the original frequency must have been lower than 440 Hz. Therefore, the initial frequency of her violin was 437 Hz.
04

Determining the Action to Tune the Violin

Since the original frequency of the violin string was lower than 440 Hz (437 Hz), to match the concert \( A \) at 440 Hz, the violinist should increase her string's frequency. Thus, she should tighten the string further.

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Key Concepts

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

Sound Frequency
Sound frequency refers to the number of sound wave cycles passing a fixed point per second. The unit of frequency is Hertz (Hz), where 1 Hz equals 1 cycle per second.
Higher frequencies are perceived as higher pitches in sound, while lower frequencies are considered lower pitches. When you hear sounds, different frequencies can result from vibrations in the air created by
  • strings, like in instruments such as violins
  • vocal cords, such as those in humans
  • speakers, which amplify electronic sound waves
Frequency is crucial in music, as it allows musicians to tune their instruments to play specific notes reliably. In the provided exercise, the sound frequency of note A (or Concert A) is given as 440 Hz, which is a standard tuning frequency. This frequency acts as a reference for tuning many musical instruments.
Violin Tuning
Tuning a violin involves adjusting the tension of its strings to match specific pitch standards. For concert performances, the A string on the violin is typically tuned to 440 Hz.
Using beat frequencies is a common method musicians employ to achieve precise tuning. In the provided exercise, the violinist listens to a reference frequency (440 Hz) while playing her A string. She notices a beat frequency, which indicates the note played by her violin and the reference tone are slightly off.
When the beat frequency decreases or stops, it shows the violin's frequency is aligning with the reference frequency. To achieve perfect tuning, the musician tweaks the string's tension accordingly:
  • If tightening the string makes the beats faster, the original frequency is lower.
  • If loosening the string makes the beats slower, the original frequency is higher.
This careful adjustment process ensures the violin is harmonized with the reference tone, enhancing musical performance.
Interference in Waves
Interference refers to the phenomenon where two or more waves superimpose to form a resultant wave.When dealing with sound waves, this interaction can lead to variations in sound intensity, known as 'beats'.
This concept is fundamental in understanding how slightly differing wave frequencies produce beats.In the case of our exercise, beats occur due to interference between the sound wave generated by the violin and the electronically produced reference tone.The beat frequency, observed here as oscillations in loudness of the sound, is calculated as \[ f_{beat} = |f_1 - f_2| \]where \( f_1 \) and \( f_2 \) are frequencies of the two interfering waves.Thus, the violinist uses this phenomenon to identify how much to adjust the instrument's frequency:
  • If the frequency difference increases, the beats become faster.
  • If the frequency difference decreases, the beats become slower or may stop.
Mastering this concept of wave interference is important not just in music, but in numerous fields involving wave mechanics.

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Most popular questions from this chapter

Longitudinal Waves in Different Fluids. (a) A longitudinal wave propagating in a water-filled pipe has intensity \(3.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) and frequency 3400 \(\mathrm{Hz}\) . Find the amplitude \(A\) and wavelength \(\lambda\) of the wave. Water has density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) and bulk modulus \(2.18 \times 10^{9} \mathrm{Pa}\) . (b) If the pipe is filled with air at pressure \(1.00 \times 10^{5} \mathrm{Pa}\) and density \(1.20 \mathrm{kg} / \mathrm{m}^{3},\) what will be the amplitude \(A\) and wavelength \(\lambda\) of a longitudinal wave with the same intensity and frequency as in part (a)? (c) In which fluid is the amplitude larger, water or air? What is the ratio of the two amplitudes? Why is this ratio so different from 1.00\(?\)

A bat flies toward a wall, emitting a steady sound of frequency 1.70 \(\mathrm{kHz}\) . This bat hears its own sound plus the sound reflected by the wall. How fast should the bat fly in order to hear a beat frequency of 10.0 \(\mathrm{Hz}\) ?

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 \(\lambda_{n}=2 L / n\) and the frequencies are given by \(f_{n}=n v / 2 L=n f_{1},\) where \(n=1,2,3, \ldots,\) (b) Modeling it as a pipe, find the frequency of the fundamental and the first two overtones for a shower 2.50 \(\mathrm{m}\) tall. Are these frequencies audible?

BIO The Human Voice. The human vocal tract is a pipe that extends about 17 \(\mathrm{cm}\) from the lips to the vocal folds (also called "vocal cords") near the middle of your throat. The vocal folds behave rather like the reed of a clarinet, and the vocal tract acts like a stopped pipe. Estimate the first three standing-wave frequencies of the vocal tract. Use \(v=344 \mathrm{m} / \mathrm{s} .\) The answers are only an estimate, since the position of lips and tongue affects the motion of air in the vocal tract.)

BIO UItrasound and Infrasound. (a) Whale communication. Blue whales apparently communicate with each other using sound of frequency 17 \(\mathrm{Hz}\) , which can be heard nearly 1000 \(\mathrm{km}\) away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is 1531 \(\mathrm{m} / \mathrm{s} ?\) (b) Dolphin clicks. One type of sound that dolphins emit is a sharp click of wavelength 1.5 \(\mathrm{cm}\) in the ocean. What is the frequency of such clicks? (c) Dog whistles. One brand of dog whistles claims a frequency of 25 \(\mathrm{kH} z\) for its product. What is the wavelength of this sound? (d) Bats. While bats emit a wide variety of sounds, one type emits pulses of sound having a frequency between 39 \(\mathrm{kHz}\) and 78 \(\mathrm{kHz}\) . What is the range of wavelengths of this sound? (e) Sonograms. Ultrasound is used to view the interior of the body, much as \(\mathrm{x}\) rays are utilized. For sharp imagery, the wave length of the sound should be around one-fourth (or less) the size of the objects to be viewed. Approximately what frequency of sound is needed to produce a clear image of a tumor that is 1.0 \(\mathrm{mm}\) across if the speed of sound in the tissue is 1550 \(\mathrm{m} / \mathrm{s} ?\)
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