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Chapter 35: Problem 1

Two small stereo speakers \(A\) and \(B\) that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point \(P\) starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker \(B\) (Fig. E35.1). For what values of x will the sound this person hears be (a) maximally reinforced, (b) cancelled? Limit your solution to the cases where x \(\leq\) 1.50 m.

Short Answer

Expert verified
Maximally reinforced: x = 0 m. Cancelled: x \approx 1.11 m.

Step by step solution

01

Determine the Path Difference

To find when sound is maximally reinforced or cancelled, we start by determining the path difference from the point \( P \) to speakers \( A \) and \( B \). Since the person at point \( P \) is always 1.50 m from speaker \( B \), the distance from \( P \) to \( A \) is \( \sqrt{(x + 0.70)^2 + 0.98^2} \), where \( 0.70 = 1.40 / 2 \) and \( 0.98 = \sqrt{1.50^2 - 0.70^2} \).
02

Maximally Reinforced Sound (Constructive Interference)

For constructive interference, the path difference must be an integer multiple of the wavelength: \( \Delta d = n\lambda \). This gives:\[ \sqrt{(x + 0.70)^2 + 0.98^2} - 1.50 = n \times 0.34 \]Solve for \( x \) using the possible values \( n=0,1,2,... \).
03

Cancelled Sound (Destructive Interference)

For destructive interference, the path difference must be an odd half multiple of the wavelength: \( \Delta d = (n+0.5)\lambda \). This leads to:\[ \sqrt{(x + 0.70)^2 + 0.98^2} - 1.50 = (n+0.5) \times 0.34 \]Solve for \( x \) using possible values for \( n \).
04

Solve the Equations within Limits

Solve the equations from Step 2 and Step 3 while limiting solutions to \( x \leq 1.50 \) m:For constructive (maximally reinforced):1. \( n = 0 \); solve the equation.2. \( n = 1 \); solve the equation, and so forth.For destructive (cancelled):1. \( n = 0 \); solve the equation.2. \( n = 1 \); solve the equation, and so forth.

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Key Concepts

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

Constructive Interference
In physics, constructive interference occurs when two or more waves overlap and combine to form a wave with a higher amplitude. This phenomenon happens when the path difference between the waves is an integer multiple of their wavelength.
For instance, if you have two sound waves that are in phase, meaning their peaks and troughs align perfectly, they amplify each other. The formula to find this is \[ \Delta d = n\lambda \]where
  • \( \Delta d \) is the path difference,
  • \( n \) is an integer (0, 1, 2, ...), and
  • \( \lambda \) is the wavelength of the sound wave.

When you stand in certain positions where this condition is met, you hear the sound much louder because the sound waves add up to reinforce each other. Think of it as spots where the sound feels just right and becomes richer and stronger.
Destructive Interference
Destructive interference is the opposite of constructive interference, where two waves interfere in such a way that they cancel each other out, leading to a lower amplitude.
This happens when the path difference is an odd multiple of half the wavelength, which can be expressed with the formula \[ \Delta d = (n+0.5)\lambda \] Here,
  • \( \Delta d \) is the path difference,
  • \( n \) is an integer (0, 1, 2, ...), and
  • \( \lambda \) is the wavelength.

When waves meet that are out of phase, meaning one wave crests where the other troughs, they effectively reduce the sound, sometimes to the point of silence.
In practical terms, this explains why you might hear little to no sound at certain points in a room with multiple sound sources.
Path Difference
Path difference is a key concept when studying wave interference. It refers to the difference in distance that two waves travel to reach a common point.
In the context of sound waves coming from two speakers, the path difference can be the difference in distance from each speaker to the listener's position. This distance affects how the waves combine.The formulae for determining if interference is constructive or destructive depend on this path difference:
  • For constructive interference: \( \Delta d = n\lambda \)
  • For destructive interference: \( \Delta d = (n+0.5)\lambda \)

By understanding and calculating the path difference, one can predict spots where sound will be loudest or quietest, a principle that's useful in designing auditoriums or speaker systems.
Sound Waves
Sound waves are a type of mechanical wave that travels through a medium, such as air, liquid, or solid. These waves are produced by vibrating objects and travel via compressions and rarefactions.Sound waves have characteristics such as
  • Wavelength (\( \lambda \)): the distance between consecutive points of phase (e.g., crest to crest).
  • Frequency: how many wavelengths pass a point in one second.
  • Amplitude: the height of the wave, which determines loudness.

Sound waves interact with each other and the environment, leading to interference patterns like constructive and destructive interference. These interactions are critical when positioning speakers for optimal sound quality or in understanding sound acoustics in a given space.

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Most popular questions from this chapter

Two slits spaced \(0.0720 \mathrm{~mm}\) apart are \(0.800 \mathrm{~m}\) from a screen. Coherent light of wavelength \(\lambda\) passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is \(3.00 \mathrm{~mm} .\) If the intensity at the peak of the central maximum is \(0.0600 \mathrm{~W} / \mathrm{m}^{2},\) what is the intensity at points on the screen that are (a) \(2.00 \mathrm{~mm}\) and (b) \(1.50 \mathrm{~mm}\) from the center of the central maximum?

Eyeglass lenses can be coated on the \(inner\) surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index 1.432, (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength 550 nm reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

Coherent light of frequency \(6.32 \times10^{14}\) Hz passes through two thin slits and falls on a screen 85.0 cm away. You observe that the third bright fringe occurs at \(\pm\)3.11 cm on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

A uniform film of TiO\(_2\) , 1036 nm thick and having index of refraction 2.62, is spread uniformly over the surface of crown glass of refractive index 1.52. Light of wavelength 520.0 nm falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the \(minimum\) thickness of TiO\(_2\) that you must \(add\) so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wavelengths of the light in the TiO\(_2\) film.

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index 1.50. At room temperature (20.0\(^\circ\)C), this film is just thick enough for light with wavelength 582.4 nm reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170\(^\circ\)C, you find that the film cancels reflected light with wavelength 588.5 nm. What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)
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