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A loudspeaker with a diaphragm that vibrates at 960 Hz is traveling at 80.0 m/s directly toward a pair of holes in a very large wall. The speed of sound in the region is 344 m/s. Far from the wall, you observe that the sound coming through the openings first cancels at \(\pm11.4^\circ\) with respect to the direction in which the speaker is moving. (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) Openings are approximately 0.688 m apart. (b) Sound cancels near \( \pm12.5^\circ \) if the source is stationary.

Step by step solution

01

Understanding the Problem

We have a source moving towards two openings at 80.0 m/s with a sound frequency of 960 Hz, and sound travels at 344 m/s. The interference creates cancellation at angles of \( \pm11.4^\circ \). We need to find (a) the distance between the openings and (b) the angles where cancellation occurs if the source stops moving.
02

Calculate the Wavelength of the Sound

The wavelength \( \lambda \) of sound when the source is moving is found using the formula \( \lambda = \frac{v_s}{f'} \) where \( v_s = 344 \text{ m/s} \) and \( f' = f \left( \frac{v_s}{v_s - v_o} \right) \). Here \( f = 960 \text{ Hz} \) and \( v_o = 80 \text{ m/s} \). Calculate \( \lambda \).
03

Find the Angle for First Minimum

The path difference for cancellation at \( \pm11.4^\circ \) is \( \lambda/2 \) for the first minimum. We use \( d \sin \theta = \frac{\lambda}{2} \) to find \( d \), where \( d \) is the distance between the openings and \( \theta = 11.4^\circ \). Solve for \( d \).
04

Calculate Wavelength when Source is Stationary

When the source stops, the wavelength is \( \lambda = \frac{v_s}{f} \). Calculate this new wavelength using \( v_s = 344 \text{ m/s} \) and \( f = 960 \text{ Hz} \).
05

Calculate angles with stationary source

Use the formula \( d \sin \theta = \frac{\lambda}{2} \) with the stationary wavelength to solve for \( \theta \) for the first cancellation where \( d \) is the same value found in Step 3.

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Key Concepts

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

Sound Interference
When sound waves interact, they can cause a phenomenon called sound interference. This is the result of two or more sound waves overlapping, which can alter the way we perceive sound. There are two main types of interference:
  • Constructive Interference: This occurs when sound waves align in such a way that their amplitudes add together, making the sound louder.
  • Destructive Interference: In this case, the sound waves are out of phase, meaning their amplitudes can cancel each other out, resulting in diminished or completely canceled sound.
Sound interference is a crucial aspect when studying acoustics, as it can greatly affect the clarity and perception of sound at various points around a source. In the context of our exercise, the interference from the loudspeaker creates specific angles where the sound cancels, indicating a point of destructive interference.
Wave Cancellation
Wave cancellation is an interesting process where two sound waves meet. If they have equal amplitude but are exactly out of phase, they can completely cancel each other out. This results in silence at certain points. In our loudspeaker example, wave cancellation causes silence at angles of \( \pm 11.4^\circ \).

The degree of cancellation depends on how perfectly out of phase the waves are. Moreover, different frequencies and speeds can affect where and when cancellation happens. When designing room acoustics or speaker setups, understanding wave cancellation helps avoid unwanted silent zones. This means arranging speakers or adjusting frequencies to ensure the best sound delivery.

In practical terms, sound engineers can use wave cancellation concepts to shape sound in concert halls or theaters to enhance acoustics or create soundproofing.
Wavelength Calculation
Wavelength is a fundamental concept that dictates the behavior of waves, including sound waves. It is the distance between successive peaks of a wave. Calculating wavelength is essential for determining how sound behaves as it interacts with the environment. The formula used for wavelength \[ \lambda = \frac{v_s}{f'} \]is adjusted according to the source's motion. Here \(v_s\) is the speed of sound, and \(f'\) is the apparent frequency when the source is moving towards the observer. This adjustment accounts for the Doppler Effect.

The Doppler Effect describes how sound frequency changes based on the motion of the source or observer, which in turn affects wavelength. When the source moves towards an observer, the frequency increases, and the wavelength decreases. Conversely, when it moves away, the frequency decreases, and the wavelength increases.

Accurate wavelength calculations help in predicting where interference patterns like wave cancellations will occur. It's also crucial in technology utilizing sound waves, such as ultrasound imaging or sonar systems.
Source Motion Effects
The movement of a sound source significantly changes how we perceive sound. This phenomenon is largely explained by the Doppler Effect. When a source moves, it changes the frequency of sound reaching an observer:
  • Towards Observer: The frequency appears higher, and the wavelength is shorter due to compressed wave fronts.
  • Away from Observer: The frequency seems lower, with longer wavelengths due to stretched wave fronts.
In the exercise, the loudspeaker travels at 80 m/s, altering the frequency and consequently affecting the angles at which sound cancels. When the source is at rest, only the natural frequency without Doppler shift is considered.

Understanding these effects is vital, particularly in fields like astronomy, where the Doppler Effect is used to determine the movement of stars and galaxies. In everyday applications, it's noticed when a passing car changes pitch as it zooms by, demonstrating the practical significance of these principles in observing and utilizing sound.

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Most popular questions from this chapter

Monochromatic electromagnetic radiation with wavelength \(\lambda\) from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width \(a\) if the wavelength is (a) 500 nm (visible light); (b) 50.0 \(\mu\)m (infrared radiation); (c) 0.500 nm (x rays)?

Monochromatic light is at normal incidence on a plane transmission grating. The first-order maximum in the interference pattern is at an angle of 11.3\(^\circ\). What is the angular position of the fourth-order maximum?

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.350 mm wide. The diffraction pattern is observed on a screen 3.00 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?

Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0250 mm wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.50 W/m\(^2\). (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, thin object, such as a strand of hair. In that case, \(a\) is the width of the strand. From actual laboratory measurements on a human hair, it was found that when a beam of light of wavelength 632.8 nm was shone on a single strand of hair, and the diffracted light was viewed on a screen 1.25 m away, the first dark fringes on either side of the central bright spot were 5.22 cm apart. How thick was this strand of hair?

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