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An organ pipe is tuned to emit a frequency of \(196.00 \mathrm{~Hz}\). When it and the G string of a violin are sounded together, ten beats are heard in a time of exactly \(8 \mathrm{~s}\). The beats become slower as the violin string is slowly tightened. What was the original frequency of the violin string?

Short Answer

Expert verified
The original frequency of the violin string was 197.25 Hz.

Step by step solution

01

Understand Beat Frequency

Beat frequency is a phenomenon that occurs when two similar frequencies are played together, resulting in a fluctuating volume at a rate equal to the difference between the two frequencies. Here, you hear 10 beats in 8 seconds, so the beat frequency is \( \frac{10 \, ext{beats}}{8 \, ext{s}} = 1.25 \, ext{Hz} \).
02

Identify Possible Violin Frequencies

The beat frequency is the difference between the frequencies of the organ pipe and the original frequency of the violin string. Therefore, if the frequency of the pipe is \( f_1 = 196.00 \, \text{Hz} \), the possible frequencies for the violin string are \( f_{\text{violin}} = f_1 \pm \text{beat frequency} \). Thus, \( f_{\text{violin}} = 196.00 \, \text{Hz} \pm 1.25 \, \text{Hz} \), resulting in two possible frequencies: \( 197.25 \, \text{Hz} \) or \( 194.75 \, \text{Hz} \).
03

Determine Original Frequency

Since the beats slow down when the string is tightened, it indicates the frequencies are moving closer. Thus, the original frequency of the violin string must have been higher than 196.00 Hz. Between the two options, 197.25 Hz and 194.75 Hz, the original frequency must be 197.25 Hz as the loosening won't reduce the frequency from 194.75 Hz.

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Key Concepts

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

Organ Pipe Frequency
Organ pipes produce sound by air vibration within them, similar to how wind instruments work. The frequency of an organ pipe depends on its length and the speed of sound. The basic formula for an open pipe is:
\[ f = \frac{v}{2L} \]where:
  • \(f\) is the frequency of the sound produced,
  • \(v\) is the speed of sound in air, and
  • \(L\) is the length of the pipe.
A specific frequency is emitted depending on these factors, and in this case, the organ pipe is tuned to 196.00 Hz. An organ pipe will emit this frequency constantly unless altered by factors such as temperature, but primarily stays stable when undisturbed. This stable frequency allows for the use of organ pipes in creating harmonics and tuning other instruments, like a violin.
A crucial fact is that musicians often use fixed-tonality instruments like organ pipes to adjust or evaluate the accuracy of tones produced by others, such as the human voice or string instruments.
Violin String Tuning
Violin string tuning involves adjusting the tension in the strings so they vibrate at the desired frequency. This is done by tightening or loosening the strings until the correct pitch is achieved. The frequency of a string is determined by the formula:
\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where:
  • \(f\) is the frequency,
  • \(L\) is the vibrating length of the string,
  • \(T\) is the tension, and
  • \(\mu\) is the linear mass density of the string.
For a G-string on a violin, altering the tension affects the frequency. In this problem, as the violin string's tension is increased, the beat frequency decreases, indicating the string's frequency moves closer to that of the organ pipe. This is because increasing tension raises the string's frequency, helping to pinpoint the initial frequency discrepancy when a beat frequency is heard. Tuning ensures that the resultant sound is harmonious and can be confirmed using beats against a reference tone, like from an organ pipe.
Frequency Difference
Frequency difference is a crucial concept when discussing beat frequency. It occurs when two slightly different frequencies are played together, creating an interference pattern. This results in beats or rhythmic interference sounds, measured as the difference in frequency values. Mathematically, beat frequency \(f_b\) can be expressed as:
\[ f_b = |f_1 - f_2| \]where:
  • \(f_1\) and \(f_2\) are the frequencies of two different sources.
In our scenario, the beat frequency is 1.25 Hz, meaning the organ pipe and the violin string frequencies differ by this amount.
This concept allows musicians to fine-tune instruments. When tightening or loosening strings, the goal is to reduce the frequency difference by aligning the frequencies for a consistent sound. Observing the change in beat frequency helps identify which frequency was too high or too low initially, providing insight for correct tuning.
Physics Problem Solving
Solving physics problems typically involves breaking down complex situations into understandable parts. Here, it's about calculating frequency relations. Begin by defining knowns and unknowns:
  • Organ pipe frequency = 196.00 Hz
  • Heard beat frequency = 1.25 Hz
  • Possible violin frequencies are calculated as: \(196.00 \text{ Hz} \pm 1.25 \text{ Hz}\)
With beat frequency, you can determine potential tuning discrepancies. The next logical step is to deduce which violin frequency applies. Since beats decrease when tension rises, the initial violin frequency was higher than the product from the pipe, resulting in 197.25 Hz.
By applying the laws of wave interference and recognizing that sound dynamics can guide physical corrections, skilled problem-solving involves analyzing changes in sound and advocating precise adjustments until frequencies converge for a uniform sound experience. Through this approach, one hones an essential skill set, enabling the solving of complex physics problems related to sound.

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Most popular questions from this chapter

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A tuning fork having a frequency of \(400 \mathrm{~Hz}\) (shown in Fig. 23-2) is moved away from an observer and toward a flat wall with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). What is the apparent frequency \((a)\) of the unreflected sound waves coming directly to the observer, and \((b)\) of the sound waves coming to the observer after reflection? ( \(c\) ) How many beats per second are heard? Assume the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\) (a) The fork, the source, is receding from the observer in the positive direction and so we use \(+v_{s} .\) It doesn't matter what the sign associated with \(v_{o}\) is since \(v_{e}=0\). $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(400 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+0}{340 \mathrm{~m} / \mathrm{s}+2.0 \mathrm{~m} / \mathrm{s}}=397.7 \mathrm{~Hz}=398 \mathrm{~Hz} $$ The source is moving away from the observer and the frequency is properly shifted down from \(400 \mathrm{~Hz}\) to \(398 \mathrm{~Hz}\). (b) Think of the wall as a source that reflects sound of the same frequency as that which impinges upon it. The wave crests reaching the wall are closer together than normally because the fork is moving toward the wall. Therefore, the wall will appear as a stationary source emitting sound of a higher frequency than \(400 \mathrm{~Hz}\). due to the \(2.0-\mathrm{m} / \mathrm{s}\) motion of the fork. Alternatively we can think of the reflected wave as if it came from a source (the wall) moving at \(2.0 \mathrm{~m} / \mathrm{s}\) toward the observer. Hence, we enter \(-v_{x}:\) $$ f_{o}=f_{s} \frac{v \pm v_{o}}{v \mp v_{s}}=(400 \mathrm{~Hz}) \frac{340 \mathrm{~m} / \mathrm{s}+0}{340 \mathrm{~m} / \mathrm{s}-2.0 \mathrm{~m} / \mathrm{s}}=402.4 \mathrm{~Hz}=402 \mathrm{~Hz} $$ and the frequency is properly shifted up. (c) Beats per second \(=\) Difference between frequencies \(=(402.4-397.7) \mathrm{Hz}=4.7\) beats per second

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