/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 108 Two tuning forks having frequenc... [FREE SOLUTION] | 91Ó°ÊÓ

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Two tuning forks having frequencies of 460 and 464 Hz are struck simultaneously. What average frequency will you hear, and what will the beat frequency be?

Short Answer

Expert verified
The average frequency heard when the two tuning forks are struck simultaneously is 462 Hz, and the beat frequency is 4 Hz.

Step by step solution

01

Calculate the average frequency

To find the average frequency of the two tuning forks, add their frequencies together and divide by 2. The formula for finding average frequency (f_avg) is: \(f_{avg} = \frac{f_1 + f_2}{2}\) where \(f_1\) and \(f_2\) represent the frequencies of the two tuning forks. Their given frequencies are 460 Hz and 464 Hz. Using the formula, we get: \(f_{avg} = \frac{460 + 464}{2}\)
02

Solve for the average frequency

Now we can solve for the average frequency: \(f_{avg} = \frac{924}{2}\) \(f_{avg} = 462\) So, the average frequency is 462 Hz.
03

Calculate the beat frequency

To find the beat frequency, we calculate the difference between the two frequencies. The beat frequency (f_beat) formula is: \(f_{beat} = |f_1 - f_2|\) Using the given frequencies, we get: \(f_{beat} = |460 - 464|\)
04

Solve for the beat frequency

Now we can solve for the beat frequency: \(f_{beat} = |-4|\) \(f_{beat} = 4\) So, the beat frequency is 4 Hz. To summarize, when the two tuning forks are struck simultaneously, you will hear an average frequency of 462 Hz and a beat frequency of 4 Hz.

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Key Concepts

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

Average Frequency
When two sound frequencies are combined, calculating the average frequency is a straightforward process. The average frequency gives us a midpoint between the two frequencies we're evaluating.
This is essential in scenarios where multiple sound sources need to create a consistent auditory experience.

For instance, in the problem we've tackled, we have two tuning forks with frequencies of 460 Hz and 464 Hz.
The formula used to determine the average frequency, \(f_{\text{avg}}\), is:\[f_{\text{avg}} = \frac{f_1 + f_2}{2}\]where \(f_1\) and \(f_2\) are the frequencies of the sound sources.
In this case:
  • Add the two frequencies: 460 + 464 = 924 Hz
  • Divide by 2: \(f_{\text{avg}} = \frac{924}{2} = 462\, \text{Hz}\)
This average of 462 Hz represents the central frequency you would perceive if the sound waves from the tuning forks were overlapping and continuous.
Tuning Forks
Tuning forks are fascinating tools primarily used to create a specific pitch or frequency for musical tuning. They consist of a metal U-shaped bar with a handle, which when struck, produces a pure tone.

This tone is a result of the vibrating metal emitting sound waves at a consistent frequency. The nature of a tuning fork ensures that the oscillations form a very precise and stable frequency, free from any unpleasant noise or distortion.

Tuning forks have been a staple in music education and performance because they offer accuracy for tuning instruments.
By providing a steady reference pitch, musicians can tune their instruments to match the tuning fork's frequency.
In the exercise we're exploring, tuning forks of 460 Hz and 464 Hz are used to show how differences in frequency can produce phenomena like beats and average frequencies, leading to richer understanding and auditory experiences.
Frequency Calculation
Frequency calculations are crucial for understanding how sound waves interact.
These calculations help us understand complex acoustic phenomena by providing baseline measurements for sound analysis.

In particular, calculating the frequency differences between two sources can reveal insightful interactions known as beats.
The beat frequency, calculated using the formula \(f_{\text{beat}} = |f_1 - f_2|\), provides the perceived oscillation rate of volume as the sound waves interfere with one another.
  • In the given problem, \(|460 - 464| = 4\, \text{Hz}\), meaning you hear 4 beats per second.
The beat frequency helps sound engineers and musicians understand and manage the auditory experience.
It can also act as a guide while tuning instruments, indicating if two notes are harmonically synchronized.

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Most popular questions from this chapter

What are the first three overtones of a bassoon that has a fundamental frequency of \(90.0 \mathrm{Hz}\) ? It is open at both ends. (The overtones of a real bassoon are more complex than this example, because its double reed makes it act more like a tube closed at one end.)

A physicist at a fireworks display times the lag between seeing an explosion and hearing its sound, and finds it to be 0.400 s. (a) How far away is the explosion if air temperature is \(24.0^{\circ} \mathrm{C}\) and if you neglect the time taken for light to reach the physicist? (b) Calculate the distance to the explosion taking the speed of light into account. Note that this distance is negligibly greater.

A sound wave of a frequency of \(2.00 \mathrm{kHz}\) is produced by a string oscillating in the \(n=6\) mode. The linear mass density of the string is \(\mu=0.0065 \mathrm{kg} / \mathrm{m}\) and the length of the string is \(1.50 \mathrm{m}\). What is the tension in the string?

The middle C hammer of a piano hits two strings, producing beats of \(1.50 \mathrm{Hz}\). One of the strings is tuned to 260.00 Hz. What frequencies could the other string have?

A 512-Hz tuning fork is struck and placed next to a tube with a movable piston, creating a tube with a variable length. The piston is slid down the pipe and resonance is reached when the piston is \(115.50 \mathrm{cm}\) from the open end. The next resonance is reached when the piston is \(82.50 \mathrm{cm}\) from the open end. (a) What is the speed of sound in the tube? (b) How far from the open end will the piston cause the next mode of resonance?

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