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Chapter 18: Problem 15

Two speakers are driven in phase 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 two speakers where relative minima of sound pressure amplitude would be expected. (Use \(v=343 \mathrm{m} / \mathrm{s} . )\)

Short Answer

Expert verified
To locate the points of relative minima, solve equations of the form |2d - 1.25| = (2n + 1) * 0.42875 / 2 for d, where n is a non-negative integer, and d must be within 0 to 1.25 meters.

Step by step solution

01

- Understanding the problem

We need to find the points between two speakers where the sound waves interfere destructively, resulting in relative minima of sound pressure amplitude. This happens at points where the path difference between the waves from the two speakers is an odd multiple of half wavelengths. The sound waves are in phase and have a frequency of 800 Hz.
02

- Calculating the wavelength

The wavelength (\( \text{位} \) ) of sound can be found using the relationship between speed (v), frequency (f) and wavelength: \( v = f \times \text{位} \). We can rearrange to solve for wavelength: \( \text{位} = \frac{v}{f} \).
03

- Applying values to find the wavelength

Now we plug in the values: \( v = 343 \text{ m/s} \) and \( f = 800 \text{ Hz} \). Thus, \( \text{位} = \frac{343 \text{ m/s}}{800 \text{ Hz}} = 0.42875 \text{ m} \).
04

- Determining conditions for destructive interference

Destructive interference occurs when the path difference is an odd multiple of half the wavelength (\( \text{m}\text{位} / 2 \) where m is an odd integer). We want to find the points along the line that satisfy this condition.
05

- Locating the points of minimum pressure

Let d be the distance from one speaker to a point of destructive interference. Since the speakers are 1.25 m apart, the path difference is the distance from the point to one speaker minus the distance from the point to the other speaker. Expressed as an equation for the nth minima, it is: \( |2d - 1.25| = n \times 0.42875 / 2 \) where n is an odd integer (1,3,5,...). We solve this equation to find d for different values of n.
06

- Writing the equation for locations of minima

Since the speakers are facing each other, d must be less than the distance between the speakers, which is 1.25 m. We set up the equation for d as \( |2d - 1.25| = (2n + 1) \times 0.42875 / 2 \), where n is an integer starting from 0. Then, we separately solve the equation for when the expression inside the absolute value is positive and negative, as the absolute value signifies that both cases should be considered.
07

- Finding the points using the equation

Solve the equations \( 2d - 1.25 = (2n + 1) \times 0.42875 / 2 \) and \( -(2d - 1.25) = (2n + 1) \times 0.42875 / 2 \) for different n and ensure d is between 0 and 1.25 m. Each valid solution for d represents a point of destructive interference, i.e., a relative minimum of sound pressure amplitude.

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Key Concepts

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

Wave Interference
When two or more sound waves meet, they interact in a phenomenon known as wave interference. This interaction can result in waves adding up to create a larger amplitude, known as constructive interference, or canceling each other out to some degree, which is known as destructive interference.

Picture two ripples on a pond colliding; at some points, the ripples will merge to make a bigger wave, while at other points, they will cancel each other leaving calm water. This is similar to what happens with sound waves. Since sound waves are pressure waves, during destructive interference, the peak pressure of one wave aligns with the trough of another, resulting in a decrease in the overall sound pressure amplitude at that point 鈥� this is the 'relative minima' referred to in the exercise.
Sound Wave Frequency
The frequency of a sound wave represents how often the particles of the medium through which the sound is transmitted vibrate back and forth each second. It's measured in hertz (Hz), and human ears can mostly hear sounds between 20 Hz to 20,000 Hz. A higher frequency corresponds to a higher pitched sound.

In the provided exercise, the frequency of the sound waves emitted by the two speakers is 800 Hz, which means the air particles vibrate 800 times per second when the sound is passing through.
Sound Pressure Amplitude
Sound pressure amplitude can be thought of as the 'strength' or 'loudness' of a sound wave at a given point and is determined by the wave's amplitude. This pressure fluctuation is what your eardrum detects and translates into sound.

In terms of destructive interference, sound pressure amplitude is crucial because it helps determine where we will perceive a decrease in volume. The points of destructive interference result in minima of sound pressure amplitude, meaning these are the points where the loudness is significantly reduced due to the overlapping of the sound waves in opposite phases.
Wavelength Calculation
The wavelength is the physical distance over which a wave's shape repeats. It is related to both the speed of sound in the medium and the frequency of the sound wave. The equation \( \lambda = \frac{v}{f} \) is used to calculate the wavelength, where \( \lambda \) is the wavelength, \( v \) is the speed of sound, and \( f \) is the frequency.

For instance, with a sound speed of 343 m/s and a frequency of 800 Hz, the wavelength is found to be about 0.429 meters. Knowing the wavelength helps in calculating the points of destructive interference by using multiples of half this value to find the points where sound waves from the speakers will meet out of phase.

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Most popular questions from this chapter

Two sinusoidal waves combining in a medium are described by the wave functions $$y_{1}=(3.0 \mathrm{cm}) \sin \pi(x+0.60 t)$$ and $$y_{2}=(3.0 \mathrm{cm}) \sin \pi(x-0.60 t)$$ where \(x\) is in centimeters and \(t\) is in seconds. Determine the maximum transverse position of an element of the medium at (a) \(x=0.250 \mathrm{cm},\) (b) \(x=0.500 \mathrm{cm},\) and (c) \(x=1.50 \mathrm{cm} .\) (d) Find the three smallest values of \(x\) corresponding to antinodes.

On a marimba (Fig. Pl8.57), the wooden bar that sounds a tone when struck vibrates in a transverse standing wave having three antinodes and two nodes. The lowest frequency note is 87.0 \(\mathrm{Hz}\) , produced by a bar 40.0 \(\mathrm{cm}\) long. (a) Find the speed of transverse waves on the bar. (b) A resonant pipe suspended vertically below the center of the bar enhances the loudness of the emitted sound. If the pipe is open at the top end only and the speed of sound in air is 340 \(\mathrm{m} / \mathrm{s}\) , what is the length of the pipe required to resonate with the bar in part (a)?

When beats occur at a rate higher than about 20 per second, they are not heard individually but rather as a steady hum, called a combination tone. The player of a typical pipe organ can press a single key and make the organ produce sound with different fundamental frequencies. She can select and pull out different stops to make the same key for the note \(\mathrm{C}\) produce sound at the following frequencies: 65.4 Hz from a so-called eight-foot pipe; \(2 \times 65.4=\) 131 \(\mathrm{Hz}\) from a four-foot pipe; \(3 \times 65.4=196 \mathrm{Hz}\) from a two-and-two-thirds-foot pipe; \(4 \times 65.4=262 \mathrm{Hz}\) from a two-foot pipe; or any combination of these. With notes at low frequencies, she obtains sound with the richest quality by pulling out all the stops. When an air leak develops in one of the pipes, that pipe cannot be used. If a leak occurs in an eight-foot pipe, playing a combination of other pipes can create the sensation of sound at the frequency that the eight-foot pipe would produce. Which sets of stops, among those listed, could be pulled out to do this?

Two waves that set up a standing wave in a long string are given by the wave functions $$y_{1}=A \sin (k x-\omega t+\phi) \quad\( and \)\quad y_{2}=A \sin (k x+\omega t)$$ Show (a) that the addition of the arbitrary phase constant \(\phi\) changes only the position of the nodes and, in particular, (b) that the distance between nodes is still one half the wavelength.

Review problem. A series of pulses, each of amplitude \(0.150 \mathrm{m},\) is sent down a string that is attached to a post at one end. The pulses are reflected at the post and travel back along the string without loss of amplitude. What is the net displacement at a point on the string where two pulses are crossing, (a) if the string is rigidly attached to the post? (b) if the end at which reflection occurs is free to slide up and down?
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