/*! 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 54 You have five tuning forks that ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You have five tuning forks that oscillate at close but different resonant frequencies. What are the (a) maximum and (b) minimum number of different beat frequencies you can produce by sounding the forks two at a time, depending on how the resonant frequencies differ?

Short Answer

Expert verified
The maximum is 10 and the minimum is 1.

Step by step solution

01

Understanding Beat Frequency

When two tuning forks are sounded together, the beat frequency produced is equal to the absolute difference in their individual resonant frequencies. That is, for two tuning forks with frequencies \( f_1 \) and \( f_2 \), the beat frequency is \( |f_1 - f_2| \). This means that by pairing different tuning forks together, we can obtain different beat frequencies.
02

Calculating Number of Pairs

With five tuning forks, we can calculate the number of unique pairs using the combination formula \( C(n, k) \), where \( n \) is the total number of forks and \( k = 2 \) is the number of forks per pair. Thus, the number of pairs is \( C(5, 2) = \frac{5!}{2!(5-2)!} = 10 \). Therefore, there are 10 different pairs of forks that can be created.
03

Maximum Number of Different Beat Frequencies

For the maximum number of different beat frequencies, assume each of these pairs produces a unique beat frequency. Since all pairs are different, there could be as many as 10 different beat frequencies, corresponding to each possible pair.
04

Minimum Number of Different Beat Frequencies

For the minimum number of different beat frequencies, assume all pairs produce the same beat frequency. This happens when the frequency difference between each pair is equal, e.g., all forks have frequencies spaced evenly apart. In such a perfectly symmetrical case, all the pairs would yield the same beat frequency. Hence, there would be just 1 beat frequency.

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.

Resonant Frequencies
At the heart of understanding tuning forks and the beat frequencies they produce lies the concept of resonant frequencies. Each tuning fork is designed to vibrate at a particular frequency, known as its resonant frequency. This frequency is the natural oscillation rate of the tuning fork when it is struck. Unlike random sounds, resonant frequencies are precise and predictable.
Resonant frequencies are significant because when different tuning forks with slightly different frequencies are played together, they interact to create interesting sound patterns. If you have ever heard a warbling or "wah-wah" sound when two musical notes are slightly out of tune with each other, you've experienced beat frequency. This phenomenon is not just a scientific curiosity, but also finds applications in tuning musical instruments and studying sound waves.
Tuning Forks
A tuning fork is a simple yet effective tool used for tuning musical instruments, creating a fixed pitch, or in scientific applications to demonstrate sound waves. It consists of a handle and two prongs (tines) which vibrate to produce sound at a specific pitch.
Each tuning fork produces a pure musical note or tone when struck. This tone is represented by its fundamental or resonant frequency. In experiments and musical tuning, multiple tuning forks can be used in combination to explore the beat frequencies generated by their interactions. This interaction helps understand how sound waves add together or cancel each other, based on their frequencies.
If you have multiple tuning forks with close but different frequencies, you can explore how their individual sounds combine into diverse auditory experiences. This is useful when trying to understand how harmony works in music or how sound engineers mix multiple sound sources.
Number of Pairs
Pairing the tuning forks becomes essential when studying how different resonant frequencies add up to generate new frequencies, such as beat frequencies. With five tuning forks, you can create pairs and examine each pair's resulting sound.
To determine all possible pairs of tuning forks, we use combinatorics. Combinatorics allows us to calculate the number of possible distinct pairings from a set of items. Specifically, if you have five tuning forks, you can use the combination formula. The formula is denoted as \( C(n, k) \), where \( n \) is the total number of items, and \( k \) is the number of items per group. In this case, \( n = 5 \) and \( k = 2 \). Therefore, the calculation \( C(5, 2) \) equals 10. Hence, you can make 10 different pairs from the five tuning forks.
These pairings help in experimenting with and observing the phenomenon of beat frequencies. By going through all these pairs, you can practically see how slight changes in sounds and frequencies create different beat effects.
Combination Formula
The combination formula \( C(n, k) \) is a vital tool when trying to determine the number of ways to group objects together. It is used to compute the number of ways to choose \( k \) items from \( n \) without regard to order. The formula is given by \[C(n, k) = \frac{n!}{k!(n-k)!}\]In practical terms, this formula can be likened to picking teams or pairs from a group, similar to our tuning fork problem.
In our context, \( n \) is the number of tuning forks available (5), and \( k \) is the number of forks per pair (2). Plugging these values into the formula, we get \[C(5, 2) = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10\]This result indicates that there are 10 distinct ways to pair up the tuning forks. The combination formula helps in strategic planning and experiments where the order of pairing is irrelevant but enumerating all possibilities is essential. By understanding how to use this formula, learners can tackle many other problems involving selection and arrangement of groups.

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

Ultrasound, which consists ultrasour of sound waves with frequencies above the human audible range, can Artery be used to produce an image of the interior of a human body. Moreover, ultrasound can be used to measure Figure 17-47 Problem \(83 .\) the speed of the blood in the body; it does so by comparing the frequency of the ultrasound sent into the body with the frequency of the ultrasound reflected back to the body's surface by the blood. As the blood pulses, this detected frequency varies. Suppose that an ultrasound image of the arm of a patient shows an artery that is angled at \(\theta=20^{\circ}\) to the ultrasound's line of travel (Fig. 17-47). Suppose also that the frequency of the ultrasound reflected by the blood in the artery is increased by a maximum of \(5495 \mathrm{~Hz}\) from the original ultrasound frequency of \(5.000000 \mathrm{MHz} .\) (a) In Fig. \(17-47\), is the direction of the blood flow rightward of leftward? (b) The speed of sound in the human arm is \(1540 \mathrm{~m} / \mathrm{s}\). What is the maximum speed of the blood? (Hint: The Doppler effect is caused by the component of the blood's velocity along the ultrasound's direction of travel.) (c) If angle \(\theta\) were greater, would the reflected frequency be greater or less?

A tube \(1.20 \mathrm{~m}\) long is closed at one end. A stretched wire is placed near the open end. The wire is \(0.330 \mathrm{~m}\) long and has a mass of \(9.60 \mathrm{~g}\). It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

In pipe \(A\), the ratio of a particular harmonic frequency to the next lower harmonic frequency is \(1.2 .\) In pipe \(B\), the ratio of a particular harmonic frequency to the next lower harmonic frequency is 1.4. How many open ends are in (a) pipe \(A\) and (b) pipe \(B\) ?

A violin string \(30.0 \mathrm{~cm}\) long with linear density \(0.650 \mathrm{~g} / \mathrm{m}\) is placed near a loudspeaker that is fed by an audio oscillator of variable frequency. It is found that the string is set into oscillation only at the frequencies 880 and \(1320 \mathrm{~Hz}\) as the frequency of the oscillator is varied over the range \(500-1500 \mathrm{~Hz}\). What is the tension in the string?

When the door of the Chapel of the Mausoleum in Hamilton, Scotland, is slammed shut, the last echo heard by someone standing just inside the door reportedly comes 15 s later. (a) If that echo were due to a single reflection off a wall opposite the door, how far from the door is the wall? (b) If, instead, the wall is \(25.7 \mathrm{~m}\) away, how many reflections (back and forth) occur?

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.