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Chapter 14: Problem 41

Two speakers are driven by a common oscillator at \(800 \mathrm{~Hz}\) and face each other at a distance of \(1.25 \mathrm{~m}\). Locate the points along a line joining the speakers where relative min-ima of the amplitude of the pressure would be expected. \((\) Use \(v=343 \mathrm{~m} / \mathrm{s} .)\)

Short Answer

Expert verified
The points along the line joining the speakers where relative minima of the amplitude of the pressure would be expected are at 0.214375 m and 0.643125 m

Step by step solution

01

Find the Wavelength

First, we need to find the wavelength of the sound. According to the formula \(v = f \cdot \lambda\), where \(v\) is the speed of sound, \(f\) is the frequency and \(\lambda\) is the wavelength. We can rearrange it as \(\lambda = \frac{v}{f}\). Substituting for \(v = 343 m/s\) and \(f = 800 Hz\), we find that \(\lambda = \frac{343}{800} = 0.42875 m\).
02

Locate the points of constructive and destructive interference

Destructive interference (area of minimum pressure amplitude) will occur at distances that are multiples of half the wavelength from each speaker along the line joining them. So the points where this happens will be \(\frac{\lambda}{2}\), \( \frac{3\lambda}{2}\), \( \frac{5\lambda}{2}\) and so on, up to the point where the distance doesn't exceed the distance between the speakers (which is 1.25 m). Given that \(\lambda = 0.42875 m\), these minimal points will be at \(0.214375 m\), \(0.643125 m\) etc along the line joining the speakers.
03

Identify valid points

We then need to identify the points that fall within the distance between the speakers (1.25 m). We find that the valid points of minimum pressure amplitude need to be less than the total distance between the speakers. So we find the valid points are \(0.214375 m, 0.643125 m \)

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Key Concepts

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

Wavelength Calculation
When working with sound waves, calculating the wavelength is a crucial first step. The wavelength (\(\lambda\)) is the distance over which the wave's shape repeats. For sound, this involves the interplay of the speed of sound (\(v\)) and the frequency (\(f\)).

The basic formula relating these quantities is:
  • Speed of sound formula: \(v = f \cdot \lambda\), where \(v = 343\; \text{m/s}\) in air at room temperature.
To calculate the wavelength, we rearrange the formula as:
  • Wavelength formula: \(\lambda = \frac{v}{f}\)
For the problem at hand, with a frequency of \(800\; \text{Hz}\), the calculation would be:
  • \(\lambda = \frac{343}{800} = 0.42875\; \text{m}\)
This means the sound wave repeats every \(0.42875 \; \text{meters}\) as it travels.
Constructive and Destructive Interference
Sound waves can interfere with each other, creating regions of maximum and minimum pressure amplitude. Constructive and destructive interference are two key concepts that describe this phenomenon.

Interference occurs when two sound waves meet, and it can be either:
  • Constructive: When waves align in phase, their amplitudes add together, creating louder regions.
  • Destructive: When waves are out of phase, their amplitudes reduce, resulting in quieter regions (or nodes).
In the context of the given exercise, we focus on destructive interference, where minimal pressure amplitude happens at distances that are multiples of half the wavelength.

Given our calculated wavelength (\(\lambda\)) of \(0.42875 \; \text{m}\), the points of destructive interference occur at:
  • \(\frac{\lambda}{2} = 0.214375\; \text{m}\)
  • \(\frac{3\lambda}{2} = 0.643125\; \text{m}\)
These are the intervals where the sound wave cancels itself, causing a drop in sound level along the line between the speakers.
Sound Frequency
Sound frequency (\(f\)), measured in Hertz (Hz), is a fundamental concept that characterizes the pitch of sound. It is defined as the number of vibrations or cycles per second of a waveform. Higher frequencies correspond to higher pitches, while lower frequencies correspond to deeper pitches.

In our scenario, the sound frequency is \(800\; \text{Hz}\). This frequency is typical of a tone that is relatively high-pitched, perceivable by the human ear.
  • The frequency determines how quickly the waves oscillate.
  • It plays a crucial role in both the wavelength calculation and interference patterns.
The importance of understanding frequency lies in its ability to influence how we hear and interact with sound, as it directly links to other properties like the speed of sound and wavelength. In exercises dealing with two or more interfering waves, knowing the frequency enables precise calculation of both constructive and destructive interference points.

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Most popular questions from this chapter

The windpipe of a typical whooping crane is about \(5.0 \mathrm{ft}\). long. What is the lowest resonant frequency of this pipe. assuming it is closed at one cnd? Assume a temperature of \(37^{\circ} \mathrm{C}\).

A trumpet creates a sound intensity level of \(1.15 \times\) \(10^{2} \mathrm{~dB}\) at a distance of \(1.00 \mathrm{~m}\). (a) What is the sound intensity of a trumpet at this distance? (b) What is the sound intensity of five trumpets at this distance? (c) Find the sound intensity of five trumpets at the location of the first row of an audience, \(8.00 \mathrm{~m}\) away, assuming, for simplicity, the sound energy propagates uniformly in all directions. (d) Calculate the decibel level of the five trumpets in the first row. (e) If the trumpets are being plaved in an outdoor auditorium, how far away, in theory, can their combined sound be heard? (f) In practice such a sound could not be heard once the listener was \(2-3 \mathrm{~km}\) away. Why can't the sound be heard at the distance found in part (e)? Hint: In a very quiet room the ambient sound intensity level is about \(30 \mathrm{~dB}\).

Two adjacent nahural frequencies of an organ pipe arc found to be \(550 \mathrm{H}_{2}\) and \(650 \mathrm{~Hz}\). Calculate the fundamenual frequency and length of this pipe. (Use \(v=340 \mathrm{~m} / \mathrm{s}\).) Determine whether the pipe is open at both ends or open at only one end.

A tumnel under a river is \(2.00 \mathrm{~km}\) long. (a) At what frequencies can the air in the tunnel resonate? (b) Explain whether it would be good to make a rule against blowing your car horn when you are in the tunnel.

Expectant parents are thrilled to hear their unborn baby's heartbeat, revealed by an ultrasonic motion detector. Suppose the fetus's ventricular wall moves in simple harmonic motion with amplitude \(1.80 \mathrm{~mm}\) and frequency 115 per minute, (a) Find the maximum linear speed of the heart wall. Suppose the motion detector in contact with the mother's abdomen produces sound at precisely \(2 \mathrm{MH} z\), which travels through tissue at \(1.50 \mathrm{~km} / \mathrm{s}\). (b) Find the maximum frequency at which sound arrives at the wall of the baby's heart. (c) Find the maximum frequency at which reflected sound is received by the motion detector. (By electronically "listening" for echoes at a frequency different from the broadcast frequency, the motion detector can produce beeps of audible sound in synchrony with the fetal heartbeat.)
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