Problem 39
Question
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?
Step-by-Step Solution
Verified Answer
(a) 437 Hz; (b) Tighten the string.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
4Step 4: 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.
Key Concepts
Sound FrequencyViolin TuningInterference in Waves
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
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
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:
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.
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:
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.
Other exercises in this chapter
Problem 36
Two loudspeakers, \(A\) and \(B,\) are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 17
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Two organ pipes, open at ond but closed at the other, are each 1.14 m long. One is now lengthened by 2.00 \(\mathrm{cm} .\) Find the frequency of the beat they
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Two organ pipes, open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2.00 \(\mathrm{cm} .\) Find the frequency of the beat t
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