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Chapter 21: Problem 60

Two loudspeakers emit sound waves of the same frequency along the \(x\) -axis. The amplitude of each wave is \(a\). The sound intensity is minimum when speaker 2 is \(10 \mathrm{cm}\) behind speaker \(1 .\) The intensity increases as speaker 2 is moved forward and first reaches maximum, with amplitude \(2 a\), when it is \(30 \mathrm{cm}\) in front of speaker \(1 .\) What is a. The wavelength of the sound? b. The phase difference between the two loudspeakers? c. The amplitude of the sound (as a multiple of \(a\) ) if the speak- ers are placed side by side?

Short Answer

Expert verified
a. The wavelength of the sound is \(20 \,cm\). b. The phase difference between the two loudspeakers is \(3蟺\) radian. c. The amplitude of the sound, if the speakers are placed side by side, is \(2a\)

Step by step solution

01

Determine the Condition for Destructive Interference

The note implies destructive interference occurred when the second speaker is \(10 \,cm\) behind the first speaker. Destructive interference occurs when the path difference between two waves is the half-integer multiple of the wavelength. In this case, it is \(位/2\), \(3位/2\), \(5位/2\), etc. However, from the given, the difference in position between speakers 1 and 2 which caused destructive interference is \(10 \,cm = 位/2\). From this, we can conclude that the wavelength \(位 = 2*10 \,cm = 20 \,cm\)
02

Calculate the Phase Difference

The phase difference between two speakers can be determined by understanding the conditions that cause maximum or minimum intensity. As we have the wavelength, we know that every \(位\) or whole number multiple of \(位\) will be a point of constructive interference, whereas every \(位/2\) or odd multiple of \(位/2\) will be a point of destructive interference. The distance that speaker is moved for constructive interference is \(30 \,cm\), which is \(1.5 * 位\). Therefore, the phase difference \(蠁 = 1.5 * 2蟺 = 3蟺\) radian.
03

Find the Amplitude when Speakers are Side by Side

When the two speakers are side by side, they are essentially at the same position. The path difference is zero, so the two waves interfere constructively. The resultant amplitude of the superposed waves is twice the amplitude of individual waves. So, the amplitude of sound, when the speakers are side-by-side, is \(2a\)

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Key Concepts

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

Destructive Interference
When two sound waves with the same frequency and amplitude, such as those emitted from loudspeakers, meet, they can cancel each other out if their crests and troughs align oppositely. This cancellation phenomenon is known as destructive interference. It occurs when the path difference between the waves is an odd number of half wavelengths, i.e., \( \lambda/2 \), \( 3\lambda/2 \), etc. In the context of the exercise, destructive interference happens when one loudspeaker is behind the other by 10 cm, indicating the path difference is \( \lambda/2 \). This allows us to calculate the wavelength of the sound wave, as it is twice the distance causing the interference, leading to a wavelength of 20 cm.
Constructive Interference
Conversely, when two sound waves meet in such a way that their crests and troughs are perfectly aligned, they reinforce each other鈥攖his is known as constructive interference. It results in a sound wave with greater amplitude than the individual waves. This happens when the path difference is a whole number of wavelengths, i.e., \( \lambda \), \( 2\lambda \), etc. In our exercise, we know that constructive interference occurs when one speaker is moved to be 30 cm in front of the other, which corresponds to a path difference of 1.5 wavelengths, leading to an amplified sound wave with double the amplitude of the individual waves.
Wavelength Calculation
The wavelength of a sound wave is the distance between two consecutive points that are in phase鈥攎eaning they are at the same point in their cycle of movement, such as two adjacent crests. To calculate the wavelength in scenarios involving interference, you can use the information about how changes in the position of one source relative to another affect the combined sound intensity. The exercise provided a clear case where a movement of 10 cm leads to destructive interference, inferring that this distance is half the wavelength. Doubling that distance gives the full wavelength, which in this example is 20 cm.
Phase Difference
The phase difference between two waves refers to the amount by which one wave is shifted relative to another. It's measured in radians and informs us about how in- or out-of-sync the waves are with each other. A phase difference of 0 or a multiple of \(2\pi \) indicates that the waves are perfectly in phase鈥攃ontributing to constructive interference. In contrast, a phase difference of \(\pi \), or an odd multiple of \(\pi\), signifies destructive interference. From the solution provided, we calculate the phase difference between the two loudspeakers to be \(3\pi \) radians when the displacement is 30 cm, correlating to 1.5 times the wavelength (\(1.5\lambda\)). Consequently, this phase difference leads us to a point of constructive interference with maximum intensity and amplitude.

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Most popular questions from this chapter

A soap bubble is essentially a very thin film of water \((n=\) 1.33) surrounded by air. The colors that you see in soap bubbles are produced by interference, much like the colors of dichroic glass. a. Derive an expression for the wavelengths \(\lambda_{\mathrm{c}}\) for which constructive interference causes a strong reflection from a soap bubble of thickness \(d\) Hint: Think about the reflection phase shifts at both boundaries. b. What visible wavelengths of light are strongly reflected from a 390 -nm- thick soap bubble? What color would such a soap bubble appear to be?

Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note \(A\) should have a frequency of \(440 \mathrm{Hz}\) and the note \(\mathrm{E}\) should be at \(659 \mathrm{Hz}\) a. What is the frequency difference between the third harmonic of the A and the second harmonic of the E? b. A tuner first tunes the A string very precisely by matching it to a \(440 \mathrm{Hz}\) tuning fork. She then strikes the A and E strings simultancously and listens for beats between the harmonics. What beat frequency indicates that the B string is properly tuned? c. The tuner starts with the tension in the B string a little low, then tightens it. What is the frequency of the B string when she hears four beats per second?

An open-open organ pipe is \(78.0 \mathrm{cm}\) long. An open-closed pipe has a fundamental frequency equal to the third harmonic of the open-open pipe. How long is the open-closed pipe?

Il Example 21.10 showed that a 92 -nm-thick coating of \(\mathrm{MgF}_{2}\) \((n=1.39)\) on glass acts as an antireflection coating for light with a wavelength of \(510 \mathrm{nm}\). Without the coating, the intensity of reflected light is \(I_{0}=c a^{2},\) where \(a\) is the amplitude of the reflected light wave and \(c\) is an unknown proportionality constant. a. Let \(I_{\lambda}\) be the intensity of light reflected from the coated glass at wavelength \(\lambda\). Find an expression for the ratio \(I_{\lambda} / I_{0}\) as a function of the wavelength \(\lambda\). This ratio is the reflection intensity from the coated glass relative to the reflection intensity from uncoated glass. A ratio less than 1 indicates that the coating is reducing the reflection intensity. Hint: The amplitude of the superposition of two waves depends on the phase difference between the waves. Although not entirely accurate, assume that both reflected waves have amplitude \(a\). b. Evaluate \(I_{A} / I_{0}\) at \(\lambda=400,450,500,550,600,650,\) and \(700 \mathrm{nm} .\) This spans the range of visible light. c. Draw a graph of \(I_{\lambda} / I_{0}\) versus \(\lambda\).

Two strings are adjusted to vibrate at exactly \(200 \mathrm{Hz}\). Then the tension in one string is increased slightly. Afterward, three beats per second are heard when the strings vibrate at the same time. What is the new frequency of the string that was tightened?
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