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A loudspeaker with a diaphragm that vibrates at \(960 \mathrm{~Hz}\) is traveling at \(80.0 \mathrm{~m} / \mathrm{s}\) directly toward a pair of holes in a very large wall. The speed of sound in the region is \(344 \mathrm{~m} / \mathrm{s}\). Far from the wall, you observe that the sound coming through the openings first cancels at \(\pm 11.4^{\circ}\) 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) The two openings are 1.52 meters apart. (b) If the source stopped moving, the sound would first cancel at ±13.62°

Step by step solution

01

Understand the provided information

The frequency of the vibrating speaker is 960 Hz. The speed of the speaker is 80.0 m/s while the speed of sound is 344 m/s. The observed angles at which sound cancels out are ±11.4 degrees when the speaker is moving.
02

Calculate the wave’s observed frequency using the Doppler effect

The observed frequency \(f'\) can be calculated using the formula for the Doppler effect: \(f' = f * (v + v_0) / v\), where \(f\) is the source frequency (960 Hz), \(v\) is the speed of sound (344 m/s), and \(v_0\) is the speed of the source (80 m/s). Plugging in the given values, the observed frequency is calculated as: \(f' = 960 * (344 + 80) / 344 = 1129.07 Hz\)
03

Compute the wavelength of the observed sound wave

The wavelength \(λ'\) can be calculated using the wave speed equation \(v = f * λ\), with rearranged to \(λ = v / f\). The wavelength of the observed sound wave is therefore: \(λ' = 344/1129.07 = 0.304 m\)
04

Calculate the distance between the openings (diffraction grating)

Based on the angle of interference, which we have as ±11.4 degrees, and through the formula sin(θ) = m*(λ/d) (where m=1 as we're calculating the first cancellation, λ is the wavelength and d is the separation between slits), we can express the distance \(d\) as: \(d = m*λ / sin(θ) = 1*0.304 / sin(11.4°) = 1.52 m\)
05

Compute the angles if the source stopped moving

When the source stops moving, the frequency remains as the original 960 Hz. Thus, the wavelength would be \(λ = v / f = 344 / 960 = 0.358 m\). The cancellation angles can then be found using sin(θ) = m*(λ/d): \(θ = arcsin(m*λ / d) = arcsin(1*0.358 / 1.52) = 13.62°\)

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Key Concepts

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

Physics of Sound
The physics of sound is a fascinating topic that deals with understanding how sound waves travel through different media. Sound waves are a type of mechanical wave that require a medium, like air, water, or a solid, to travel. They are created by a source, such as a loudspeaker diaphragm, vibrating at a certain frequency. This frequency, measured in hertz (Hz), determines the pitch of the sound.

The speed of sound, however, varies depending on the medium it's traveling through and the temperature of that medium. In air at room temperature, for instance, it travels at approximately 344 meters per second (m/s). It's this speed that helps us determine the wavelength of the sound wave using the formula
\( \lambda = \frac{v}{f} \)
, where
\( \lambda \)
is the wavelength,
\( v \)
is the speed of sound, and
\( f \)
is the frequency.
Wave Interference
Wave interference is the phenomenon that occurs when two or more waves meet and combine to form a new wave pattern. This can result in various outcomes, including constructive interference, where waves add together to increase amplitude, or destructive interference, where waves cancel each other out.

In the exercise, we see a case of destructive interference where the sound waves coming through two openings in a wall cancel each other at specific angles, resulting in silence at those points. This effect can be understood and calculated using the principle of differential path length and the equation
\( \sin(\theta) = \frac{m\lambda}{d} \)
, where
\( \theta \)
is the angle of cancellation,
\( m \)
is the order of cancellation,
\( \lambda \)
is the wavelength, and
\( d \)
is the distance between the openings, or slits. This equation helps us find how far apart the slits are (
\( d \)
) by using the angle of the first interference cancellation (
\( \theta \)
).
Doppler Effect
The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source. It's commonly heard when an ambulance siren changes pitch as it speeds past us, but it also significantly impacts sound and light waves.

In the context of the exercise, when the loudspeaker moves towards the wall, the frequency of the sound waves it emits appears increased to an observer in front of the speaker. A higher frequency corresponds to a higher pitch. The equation for the Doppler effect in sound is:
\( f' = f \cdot \frac{v \pm v_0}{v} \)
, where
\( f' \)
is the observed frequency,
\( f \)
is the source frequency,
\( v \)
is the speed of sound, and
\( v_0 \)
is the velocity of the source. The sign between
\( v \)
and
\( v_0 \)
depends on whether the source is moving towards or away from the observer.

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Most popular questions from this chapter

(a) What is the wavelength of light that is deviated in the first order through an angle of \(13.5^{\circ}\) by a transmission grating having 5000 slits \(/ \mathrm{cm} ?\) (b) What is the second-order deviation of this wavelength? Assume normal incidence.

When the light is passed through the bottom of the sample container, the interference maximum is observed to be at \(41^{\circ} ;\) when it is passed through the top, the corresponding maximum is at \(37^{\circ} .\) What is the best explanation for this observation? (a) The microspheres are more tightly packed at the bottom, because they tend to settle in the suspension. (b) The microspheres are more tightly packed at the top, because they tend to float to the top of the suspension. (c) The increased pressure at the bottom makes the microspheres smaller there. (d) The maximum at the bottom corresponds to \(m=2,\) whereas the maximum at the top corresponds to \(m=1\).

Laser light of wavelength \(632.8 \mathrm{nm}\) falls normally on a slit that is \(0.0250 \mathrm{~mm}\) wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is \(8.50 \mathrm{~W} / \mathrm{m}^{2}\). (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.

The Hubble Space Telescope has an aperture of \(2.4 \mathrm{~m}\) and focuses visible light \((380-750 \mathrm{nm})\). The Arecibo radio telescope in Puerto Rico is \(305 \mathrm{~m}(1000 \mathrm{ft})\) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength \(75 \mathrm{~cm}\). (a) Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon? (b) If the Hubble Space Telescope were to be converted to surveillance use, what is the highest orbit above the surface of the earth it could have and still be able to resolve the license plate (not the letters, just the plate) of a car on the ground? Assume optimal viewing conditions, so that the resolution is diffraction limited.

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 \(632.8 \mathrm{nm}\) was shone on a single strand of hair, and the diffracted light was viewed on a screen \(1.25 \mathrm{~m}\) away, the first dark fringes on either side of the central bright spot were \(5.22 \mathrm{~cm}\) apart. How thick was this strand of hair?

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