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Chapter 21: Problem 22

Two in-phase loudspeakers separated by distance \(d\) emit \(170 \mathrm{Hz}\) sound waves along the \(x\) -axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on. What are three possible values for d? Assume a sound speed of \(340 \mathrm{m} / \mathrm{s}\).

Short Answer

Expert verified
Three possible values for \(d\) are \(d = 1/2 \lambda\), \(d = 3/2 \lambda\), and \(d = 5/2 \lambda\), where \(\lambda\) is the wavelength.

Step by step solution

01

Calculate the Wavelength

First, use the formula \(v = f \lambda\) where \(v\) is the speed of sound, \(f\) is the frequency, and \(\lambda\) is the wavelength. We know that \(v = 340 m/s\) and \(f = 170 Hz\). Therefore, we can calculate \(\lambda = v / f\).
02

Determine the Possible Values for d

To achieve a destructive interference, the path difference of waves coming from both speakers would be \((n + 1/2) \lambda\), where \( n=0,1,2,3,...\). We know that this path difference equals to the distance between the loudspeakers (that is \(d\)). So, the three smallest possible values for \(d\) corresponding to \(n=0,1,2\) are \(d = 1/2 \lambda\), \(d = 3/2 \lambda\), and \(d = 5/2 \lambda\).
03

Calculate the Values for d

Lastly, substitute the value of \(\lambda\) from step 1 into the expressions from step 2 to find the values for \(d\)

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Key Concepts

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

Sound waves
Sound waves are a type of mechanical wave that travels through a medium, such as air, water, or solid materials. These waves carry energy from one point to another by causing particles within the medium to vibrate. This vibration produces a back-and-forth motion that moves outward from the source, similar to ripples on a pond.
In the context of the original exercise, sound waves are emitted from two loudspeakers. Since these speakers emit in-phase sound waves, their crests and troughs align perfectly. This means they start their oscillations at the same time. When the waves from two sources meet, they can interact in various ways, which leads us to the next important concept: interference.
Destructive interference
Destructive interference occurs when two or more waves meet and combine to produce a wave of reduced amplitude. For sound waves, this means that you end up hearing less sound, or in some cases, no sound at all, around certain points.
This can happen when the crest of one wave coincides with the trough of another. The waves essentially "cancel" each other out. In mathematical terms, destructive interference happens when the path difference between the two sound waves is an odd multiple of half wavelengths, expressed as
  • \(d = (n + 1/2) \lambda\)
  • where \(n\) is any whole number (0, 1, 2, ...).
In the exercise, this principle explains why you might not hear anything as you walk away from the speakers at certain spots, despite both being on.
Wavelength
Wavelength is the distance between consecutive crests or troughs in a wave. It is denoted by the Greek letter \(\lambda\) and is measured in meters. Wavelength is a crucial characteristic of any wave, as it determines many of the wave's properties, including its speed and the frequency at which it vibrates.
To find the wavelength \(\lambda\), you can use the formula:
  • \(\lambda = \frac{v}{f}\)
  • where \(v\) is the speed of the wave in the medium (for sound in air, approximately \(340 \text{ m/s}\))
  • and \(f\) is the frequency of the wave.
In the given exercise, the frequency is \(170 \text{ Hz}\), which allows us to calculate the wavelength and use it to determine the locations where destructive interference happens.
Frequency
Frequency refers to the number of wave cycles that pass a specific point per second. It's measured in Hertz (Hz), where 1 Hz equals one cycle per second. Frequency determines the pitch of sound: higher frequencies correspond to higher-pitched sounds, and lower frequencies to lower-pitched sounds.
The frequency of sound waves emitted by the speakers in this problem is \(170 \text{ Hz}\). This frequency plays a key role in calculating the wavelength and understanding how and where interference occurs. Since frequency is directly related to both the speed of a wave and its wavelength (as seen in the formula \(v = f \lambda \)), it is a central concept in analyzing wave behavior. Understanding this concept helps explain why at certain distances from the speakers, destructive interference leads to silence.

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Most popular questions from this chapter

Ultrasound has many medical applications, one of which is to monitor fetal heartbeats by reflecting ultrasound off a fetus in the womb. a. Consider an object moving at speed \(v_{\mathrm{o}}\) toward an at-rest source that is emitting sound waves of frequency \(f_{0} .\) Show that the reflected wave (i.c., the echo) that returns to the source has a Doppler- shifted frequency $$f_{\text {echo }}=\left(\frac{v+v_{\mathrm{o}}}{v-v_{\mathrm{o}}}\right) f_{0}$$ where \(v\) is the speed of sound in the medium. b. Suppose the object's speed is much less than the wave speed: \(v_{\mathrm{o}} \ll v .\) Then \(f_{\text {echo }} \approx f_{0},\) and a microphone that is sensitive to these frequencies will detect a beat frequency if it listens to \(f_{0}\) and \(f_{\text {echo }}\) simultaneously. Use the binomial approximation and other appropriate approximations to show that the beat frequency is \(f_{\text {beat }} \approx\left(2 v_{0} / v\right) f_{0}\) c. The reflection of \(2.40 \mathrm{MHz}\) ultrasound waves from the sur- face of a fetus's beating heart is combined with the \(2.40 \mathrm{MHz}\) wave to produce a beat frequency that reaches a maximum of \(65 \mathrm{~Hz}\). What is the maximum speed of the surface of the heart? The speed of ultrasound waves within the body is \(1540 \mathrm{~m} / \mathrm{s}\) d. Suppose the surface of the heart moves in simple harmonic motion at 90 beats/min. What is the amplitude in \(\mathrm{mm}\) of the heartbeat?

As the captain of the scientific team sent to Planet. Physics, one of your tasks is to measure \(g\). You have a long, thin wire labeled \(1.00 \mathrm{g} / \mathrm{m}\) and a \(1.25 \mathrm{kg}\) weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the \(1.25 \mathrm{kg}\) weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is \(100 \mathrm{Hz}\). Next, with the \(1.25 \mathrm{kg}\) weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 314 s to complete 100 oscillations. Pulling out your trusty calculator, you get to work. What value of \(g\) will you report back to headquarters?

A water wave is called a deep-water wave if the water's depth is more than one-quarter of the wavelength. Unlike the waves we've considered in this chapter, the speed of a deep-water wave depends on its wavelength: $$ v=\sqrt{\frac{g \lambda}{2 \pi}} $$ Longer wavelengths travel faster. Let's apply this to standing waves. Consider a diving pool that is \(5.0 \mathrm{m}\) decp and \(10.0 \mathrm{m}\) wide. Standing water waves can set up across the width of the pool. Because water sloshes up and down at the sides of the pool, the boundary conditions require antinodes at \(x=0\) and \(x=L\). Thus a standing water wave resembles a standing sound wave in an open-open tube. a. What are the wavelengths of the first three standing-wave modes for water in the pool? Do they satisfy the condition for being deep-water waves? Draw a graph of each. b. What are the wave speeds for each of these waves? c. Derive a general expression for the frequencies \(f_{m}\) of the possible standing waves. Your expression should be in terms of \(m, g,\) and \(L\) d. What are the oscillation periods of the first three standing wave modes?

A particularly beautiful note reaching your car from a rare Stradivarius violin has a wavelength of \(39.1 \mathrm{cm} .\) The room is slightly warm, so the speed of sound is \(344 \mathrm{m} / \mathrm{s}\). If the string's linear density is \(0.600 \mathrm{g} / \mathrm{m}\) and the tension is \(150 \mathrm{N}\), how long is the vibrating section of the violin string?

The lowest note on a grand piano has a frequency of \(27.5 \mathrm{Hz}\) The entire string is \(2.00 \mathrm{m}\) long and has a mass of \(400 \mathrm{g}\). The vibrating section of the string is \(1.90 \mathrm{m}\) long. What tension is needed to tune this string properly?
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