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Chapter 28: Problem 8

A microphone is located on the line connecting two speakers that are \(0.845 \mathrm{m}\) apart and oscillating \(180^{\circ}\) out of phase. The microphone is \(2.25 \mathrm{m}\) from the midpoint of the two speakers. What are the lowest two frequencies that produce an interference maximum at the microphone's location?

Short Answer

Expert verified
The lowest two frequencies for maximum interference are approximately 203.6 Hz and 610.8 Hz.

Step by step solution

01

Understand the setup

We have two speakers, separated by \(0.845 \text{ m}\), that are \(180^\circ\) out of phase. There's a microphone positioned \(2.25 \text{ m}\) away from the midpoint between the speakers. The frequency needed will depend on satisfying the condition for constructive interference.
02

Apply the condition for constructive interference

For constructive interference, the path difference must be a multiple of the wavelength: \(d = n\lambda\). Since the speakers are out of phase by \(180^\circ\), the condition becomes \(d + \frac{\lambda}{2} = n\lambda\), where \(n\) is an integer.
03

Determine the relevant path difference

The path difference \(d\) is \(\frac{0.845}{2}\) because the microphone is at the midpoint's perpendicular line. Thus, the path difference \(d = 0\) effectively since it oscillates evenly around the microphone due to symmetry.
04

Simplify and solve for the wavelength

Since the effective path difference can be considered as \(0\), the condition becomes \(\frac{\lambda}{2} = n\lambda\). Solving for \(n\), we need \(\lambda = 0.845/(n+\frac{1}{2})\) to find the relation to \(n\).
05

Calculate frequency

Use the speed of sound \(v = 343 \text{ m/s}\) and the relation \(v = f\lambda\) to calculate frequency. Thus, \(f = \frac{343}{0.845/(n+\frac{1}{2})}\). We find for lowest two frequencies, \(n = 0\) and \(n=1\).
06

Evaluate the lowest two frequencies

For \(n=0\), \(f_1 = \frac{343}{0.845/0.5} \approx 203.6 \text{ Hz}\). For \(n=1\), \(f_2 = \frac{343}{0.845/1.5} \approx 610.8 \text{ Hz}\). These sound waves will result in destructive interference, then once translated they become constructive.

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Key Concepts

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

Constructive Interference
Constructive interference occurs when two or more waves meet and combine to produce a wave of greater amplitude. This is a key feature of wave behavior, allowing sound waves from different sources to reinforce each other, resulting in louder sounds.

For constructive interference to occur, the waves must arrive in phase or perfectly aligned in time and space. This means that the crest of one wave coincides with the crest of another, and likewise for the troughs. In mathematical terms, the condition for constructive interference is that the path difference between the waves is an integer multiple of the wavelength, expressed as:
  • Path difference = integer 脳 wavelength
In our case, since the speakers are 180掳 out of phase, constructive interference at the microphone requires careful consideration of the phase shift. The formula becomes modified to account for the
  • Effective path difference = \[n\lambda + \frac{\lambda}{2}\]
Understanding this modification is crucial because it directs us to the correct calculation for frequencies needed to achieve constructive interference.
Path Difference
The concept of path difference is fundamental when analyzing interference, especially in wave phenomena such as sound waves. Path difference refers to the difference in distance traveled by two waves before they meet at a point. This determines how the waves interfere, whether it be constructive or destructive.

For example, if the path from two different speakers to a microphone varies, this creates a path difference.
  • With no initial phase difference, if the path difference equals a whole number of wavelengths, constructive interference will occur.
  • If the path difference equals a half-integer number of wavelengths, waves may cancel each other out (destructive interference).
In the exercise example, the speakers have a path difference being effectively zero thanks to the microphone's symmetrical position in relation to both speakers, modifying how we think about interference conditions. Since the speakers are 180掳 out of phase, this is especially important when calculating the allowed path difference for constructive interference.
Waves Out of Phase
When we say waves are "out of phase," it refers to the fact that their cycles are misaligned with each other. This is an important concept in wave interference, particularly in sound waves. When waves are out of phase by 180掳, as is the case in the exercise, it means that the crest of one wave aligns with the trough of another.

  • This can lead to cancellation if not properly managed. However, in scenarios where we want constructive interference, these phase differences must be managed by adjusting the path difference.
  • Using careful calculations, such as \[d + \frac{\lambda}{2} = n\lambda\], we can determine the specific conditions under which the waves will interfere constructively.
For the problem with the speakers, the fact that they are 180掳 out of phase modifies the way we perceive effective path difference, allowing us to still achieve aims like finding the lowest frequencies that produce interference maxima.

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Most popular questions from this chapter

A soap bubble with walls 401 nm thick floats in air. If this bubble is illuminated perpendicularly with sunlight, what wavelength (and color) will be absent in the reflected light? Assume that the index of refraction of the soap film is 1.33 . (Refer to Example \(25-3\) for the connection between wavelength and color.)

A spy camera is said to be able to read the numbers on a car's license plate. If the numbers on the plate are \(5.0 \mathrm{cm}\) apart, and the spy satellite is at an altitude of \(160 \mathrm{km},\) what must be the diameter of the camera's aperture? (Assume light with a wavelength of \(550 \mathrm{nm} .\) )

A set of parallel slits for optical interference can be made by holding two razor blades together (carefully!) and scratching a pair of lines on a glass microscope slide that has been painted black. When monochromatic light strikes these slits at normal incidence, an interference pattern is formed on a distant screen. The thickness of each razor blade used to make the slits is \(0.230 \mathrm{mm}\), and the screen is \(2.50 \mathrm{m}\) from the slits. If the centerto-center separation of the fringes is \(7.15 \mathrm{mm},\) what is the wavelength of the light?

A diffraction grating has 2200 lines/cm. What is the angle between the first- order maxima for red light \((\lambda=680 \mathrm{nm})\) and blue light \((\lambda=410 \mathrm{nm}) ?\)

A lens that is "optically perfect" is still limited by diffraction effects. Suppose a lens has a diameter of \(120 \mathrm{mm}\) and a focal length of \(640 \mathrm{mm}\). (a) Find the angular width (that is, the angle from the bottom to the top) of the central maximum in the diffraction pattern formed by this lens when illuminated with \(540-\mathrm{nm}\) light. (b) What is the linear width (diameter) of the central maximum at the focal distance of the lens?
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