/*! 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 23 Two out-of-tune flutes play the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two out-of-tune flutes play the same note. One produces a tone that has a frequency of \(262 \mathrm{Hz},\) while the other produces \(266 \mathrm{Hz}\). When a tuning fork is sounded together with the \(262-\mathrm{Hz}\) tone, a beat frequency of \(1 \mathrm{Hz}\) is produced. When the same tuning fork is sounded together with the \(266-\mathrm{Hz}\) tone, a beat frequency of \(3 \mathrm{Hz}\) is produced. What is the frequency of the tuning fork?

Short Answer

Expert verified
Solution not complete yet: follow step-by-step.

Step by step solution

01

Understanding Beat Frequency

When two frequencies interfere, they produce a beat frequency, which is equal to the absolute difference between the two frequencies involved. If the tuning fork frequency is \( f \), then the beat frequency with the 262 Hz tone is \( \left| f - 262 \right| = 1 \) Hz, and with the 266 Hz tone is \( \left| f - 266 \right| = 3 \) Hz.
02

Solve for Frequency with 262 Hz Tone

For the 262 Hz tone, we have:\[\left| f - 262 \right| = 1\]This gives us two possible scenarios: 1. \( f - 262 = 1 \Rightarrow f = 263 \)2. \( 262 - f = 1 \Rightarrow f = 261 \)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Beat Frequency
When two sound waves of slightly different frequencies play together, a phenomenon occurs known as beat frequency. This is experienced as a rhythmic pulsation of sound, rather than a steady tone, which arises when the two waves interfere. The beat frequency can help us identify the difference between the two notes, as it is simply the absolute difference between their frequencies. For example, if one sound has a frequency of 262 Hz and another has a frequency of 266 Hz, the beat frequency would be \( | 266 - 262 | = 4 \) Hz. This principle is handy for tuning musical instruments or determining the frequency of a tone.
Tuning Fork
A tuning fork is a simple instrument used to produce a fixed pitch. When struck, it generates a pure musical tone which is consistent and reliable, making it perfect for comparing frequencies. Tuning forks are often used in experiments or musical settings to ensure instruments are playing at the correct pitch. In this exercise, the tuning fork's frequency might not be initially known, but it can be deduced by examining how it interacts with other known frequencies, such as in our example with the flute tones. By analyzing the beat frequencies produced with each flute tone, we can back-calculate the tuning fork's own frequency.
Sound Waves
Sound waves are longitudinal waves that travel through a medium, such as air or water, by vibrating the molecules in that medium. These vibrations create areas of compression and rarefaction, which human ears interpret as sound. The frequency of a sound wave determines the pitch we hear, where higher frequencies correspond to higher pitches. In the case of musical instruments like flutes and tuning forks, these frequencies must match for harmonious sound. The combination of different frequencies leads to the creation of beats, providing useful information such as the tuning discrepancies in instruments.
Interference
Interference occurs when sound waves overlap, leading to the combination of their amplitudes either constructively or destructively. Constructive interference happens when waves align in phase, amplifying the sound. In contrast, destructive interference occurs when they are out of phase, potentially canceling each other out. This interaction is responsible for the rhythmic pulsations or beats we discussed earlier in the context of beat frequency. Through interference, musicians and technicians can determine disparities in tuning by assessing the variations in sound amplitude, allowing for precise adjustments to achieve the desired harmony. Understanding interference is crucial for interpreting wave behaviors such as beats.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two speakers, one directly behind the other, are each generating a 245-Hz sound wave. What is the smallest separation distance between the speakers that will produce destructive interference at a listener standing in front of them? The speed of sound is \(343 \mathrm{m} / \mathrm{s}\).

The A string on a string bass vibrates at a fundamental frequency of \(55.0 \mathrm{Hz}\). If the string's tension were increased by a factor of four, what would be the new fundamental frequency?

A string is fixed at both ends and is vibrating at \(130 \mathrm{Hz},\) which is its third harmonic frequency. The linear density of the string is \(5.6 \times\) \(10^{-3} \mathrm{kg} / \mathrm{m},\) and it is under a tension of \(3.3 \mathrm{N} .\) Determine the length of the string.

A tube with a cap on one end, but open at the other end, has a fundamental frequency of \(130.8 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s}\) (a) If the cap is removed, what is the new fundamental frequency of the tube? (b) How long is the tube?

Multiple-Concept Example 3 reviews the concepts that are important in this problem. The entrance to a large lecture room consists of two side-by-side doors, one hinged on the left and the other hinged on the right. Each door is \(0.700 \mathrm{m}\) wide. Sound of frequency \(607 \mathrm{Hz}\) is coming through the entrance from within the room. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the diffraction angle \(\theta\) of the sound after it passes through the doorway when (a) one door is open and (b) both doors are open?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.