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Chapter 36: Problem 54

A loudspeaker having a diaphragm that vibrates at 1250 \(\mathrm{Hz}\) is traveling at 80.0 \(\mathrm{m} / \mathrm{s}\) directly toward a pair of holes in a very large wall in a region for which the speed of sound is 344 \(\mathrm{m} / \mathrm{s} .\) You observe that the sound comind through the openings first cancels at \(\pm 11.4^{\circ}\) with respect to the original direction of the speaker when observed far from the wall. (a) How far apart are the two openings? (b) At what angles would the sound first cancel if the source stopped moving?

Short Answer

Expert verified
(a) 0.531 m; (b) 卤15.1掳.

Step by step solution

01

Understand the Doppler Effect

The loudspeaker is moving towards a stationary observer, so we use the Doppler effect formula to determine the frequency of the sound waves received by the observer. The formula is \( f' = \frac{f(v + v_o)}{v} \), where \( f' \) is the observed frequency, \( f \) is the source frequency (1250 Hz), \( v \) is the speed of sound (344 m/s), and \( v_o \) is the speed of the observer (80 m/s).
02

Calculate Observed Frequency

With the source moving towards the observer, the observed frequency \( f' \) is calculated as follows: \[ f' = \frac{1250(344 + 0)}{344 - 80} = \frac{1250 \times 344}{264} \]. Performing the calculation gives \( f' \approx 1626.5 \text{ Hz} \).
03

Determine Wavelength of the Observed Frequency

The wavelength \( \lambda \) is calculated using \( \lambda = \frac{v}{f'} = \frac{344}{1626.5} \). Solving this gives \( \lambda \approx 0.2115 \text{ m} \).
04

Use Interference Conditions

Sound cancels when the path difference between waves from the two openings equals an odd multiple of half of the wavelength: \( d \sin \theta = (n + \frac{1}{2})\lambda \), where \( \theta \) is the angle (11.4掳), \( n \) is the order of the cancellation (assuming primary cancellation with \( n=0 \)), and \( d \) is the distance between the openings.
05

Solve for Distance Between Holes

For \( n=0 \), the condition simplifies to \( d \sin 11.4^{\circ} = \frac{1}{2} \times 0.2115 \), leading to \( d = \frac{0.10575}{\sin 11.4^\circ} \approx 0.531 \text{ m} \).
06

Recalculate for Stationary Source

With the source stationary, frequency \( f = 1250 \text{ Hz} \) and thus \( \lambda = \frac{344}{1250} = 0.2752 \text{ m} \). Use \( d \sin \theta = (n + \frac{1}{2})\lambda \) for the primary cancellation, \( d = 0.531 \text{ m} \).
07

Find New Angles for Stationary Source

Substitute \( \lambda = 0.2752 \text{ m} \) into the equation: \( 0.531 \sin \theta = \frac{1}{2} \times 0.2752 \). Solving for \( \sin \theta \) gives \( \sin \theta \approx 0.2591 \), and the first angles are \( \theta \approx \pm 15.1^{\circ} \).

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Key Concepts

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

Sound Wave Interference
Sound wave interference occurs when two or more sound waves overlap and combine, leading to regions of increased or decreased sound intensity. This phenomenon can result in constructive interference, where the sound is louder, or destructive interference, where the sound is diminished, or even canceled. When sound waves pass through two closely spaced openings, they can interfere with each other, depending on the path difference between the waves traveling from each opening to a common point.
The path difference determines whether the waves align to cancel each other (destructive interference) or amplify each other (constructive interference). Interference plays a crucial role in numerous applications, such as in noise-canceling headphones and in the design of musical instruments. Understanding sound wave interference is essential for solving problems related to wave behaviors and their practical applications.
Wavelength Calculation
Wavelength is a critical parameter in understanding waves. It refers to the physical length of one complete wave cycle, usually measured in meters. The relationship between wavelength (\( \lambda \)), frequency (\( f \)), and the speed of sound (\( v \)) is described by the equation:
  • \[ \lambda = \frac{v}{f} \]
In this equation, the speed of sound is divided by the frequency of the sound wave to find the wavelength. When dealing with problems involving moving sound sources, as seen in the Doppler Effect, it is important to adjust the frequency based on the source's movement before calculating the wavelength. This ensures an accurate depiction of the wave dynamics involved. Accurate wavelength calculations are crucial in various technological and scientific applications, such as in acoustics and even astronomical observations.
Frequency Shift
The frequency shift, often seen as a result of the Doppler Effect, occurs when there is a change in frequency perceived by an observer, due to the relative motion between the observer and the source of the sound. When a sound source moves toward an observer, the frequency appears higher, and when it moves away, the frequency seems lower. This can be calculated using the formula for the Doppler Effect:
  • \[ f' = \frac{f(v + v_o)}{v - v_s} \]
where \( f' \) is the observed frequency, \( f \) is the original frequency of the source, \( v \) is the speed of sound in the medium, \( v_o \) is the velocity of the observer, and \( v_s \) is the velocity of the source.
The Doppler Effect is not only significant in acoustics but also in radar and medical imaging technologies, such as ultrasound. Understanding frequency shifts is essential for interpreting real-world scenarios where sound and motion interact.
Wave Cancellation Angles
Wave cancellation angles refer to the specific angles at which waves interfere destructively, leading to minimal or no sound intensity. For waves emerging from two openings, this angle is the result of the interference condition: when the path difference equals an odd multiple of half the wavelength. Mathematically, it can be expressed as:
  • \[ d \sin \theta = (n + \frac{1}{2}) \lambda \]
where \( d \) is the distance between the openings, \( \theta \) is the angle from the original line of sight, \( n \) is the integer order number, and \( \lambda \) is the wavelength.
In practical terms, knowing where waves cancel allows for engineering applications like designing concert halls and speaker setups to minimize unwanted noise or create zones of quiet. Understanding these angles helps to control sound propagation and achieve desirable acoustic environments. Thus, calculating wave cancellation angles is vital for managing sound interference in various fields.

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Most popular questions from this chapter

Intensity Pattern of \(N\) Slits. (a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda .\) Show that at a time \(t,\) the electric field at a distant point \(P\) is $$\begin{aligned} E_{P}(t)=& E_{0} \cos (k R-\omega t)+E_{0} \cos (k R-\omega t+\phi) \\\ &+E_{0} \cos (k R-\omega t+2 \phi)+\cdots \\ &+E_{0} \cos (k R-\omega t+(N-1) \phi) \end{aligned}$$ where \(E_{0}\) is the amplitude at \(P\) of the electric field due to an indi- vidual slit, \(\phi=(2 \pi d \sin \theta) / \lambda, \theta\) is the angle of the rays reaching \(P\) (as measured from the perpendicular bisector of the slit arrange- ment), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d\) . (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship $$e^{i z}=\cos z+i \sin z$$ where \(i=\sqrt{-1}\) . In this expression, cos \(z\) is the real part of the complex number \(e^{i z},\) and \(\sin z\) is its imaginary part. Show that the electric field \(E_{P}(t)\) is equal to the real part of the complex quantity $$\sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega t+n \phi)}$$ (c) Using the properties of the exponential function that \(e^{A} e^{B}=e^{(A+B)}\) and \(\left(e^{A}\right)^{n}=e^{n A},\) show that the sum in part (b) can be written as $$E_{0}\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)}\( \)\quad=E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{i[k R-\omega t+(N-1) \phi}$$ Then, using the relationship \(e^{i z}=\cos z+i \sin z,\) show that the (real) electric field at point \(P\) is $$I=I_{0}\left[\frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right]^{2}$$ where \(I_{0}\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N=2 .\) It will help to recall that \(\sin 2 A=2 \sin A \cos A .\) Explain why your result differs from Eq. \((35.10),\) the expression for the intensity in two-source interfer- ence, by a factor of \(4 .\) Hint: Is \(I_{0}\) defined in the same way in both expressions?

Light of wavelength 585 nm falls on a slit 0.0666 \(\mathrm{mm}\) wide. (a) On a very large and distant screen, how many totally dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem without calculating all the angles! (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest that \(m\) can be? (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 \(\mathrm{nm}\) ) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 \(\mathrm{nm} ) ?\)

Measuring Refractive Index. A thin slit illuminated by light of frequency \(f\) produces its first dark band at \(\pm 38.2^{\circ}\) in air. When the entire apparatus (slit, screen, and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at \(\pm 21.6^{\circ} .\) Find the refractive index of the liquid.

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.350 \(\mathrm{mm}\) wide. The diffraction pattern is observed on a screen 3.00 \(\mathrm{m}\) away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?
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