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An avalanche of sand along some rare desert sand dunes can produce a booming that is loud enough to be heard \(10 \mathrm{~km}\) away. The booming apparently results from a periodic oscillation of the sliding layer of sand \(-\) the layer's thickness expands and contracts. If the emitted frequency is \(90 \mathrm{~Hz}\), what are (a) the period of the thickness oscillation and (b) the wavelength of the sound?

Short Answer

Expert verified
(a) 0.0111 s; (b) 3.77 m

Step by step solution

01

Find the Period of Oscillation for Frequency

The period of an oscillation \( T \) is the reciprocal of the frequency \( f \). Given the frequency \( f = 90 \ \, \mathrm{Hz} \), the formula to find the period is:\[T = \frac{1}{f}\]Plug in the value of \( f \):\[T = \frac{1}{90 \, \mathrm{Hz}} \approx 0.0111 \, \mathrm{s}\]
02

Calculate the Speed of Sound

To determine the wavelength, we first need the speed of sound. In a typical desert environment, the speed of sound is approximately 340 meters per second (m/s).
03

Find the Wavelength of the Sound

Wavelength \( \lambda \) can be calculated using the formula:\[\lambda = v \cdot T\]where \( v \) is the speed of sound and \( T \) is the period calculated previously. Substituting the known values:\[\lambda = 340 \, \mathrm{m/s} \cdot 0.0111 \, \mathrm{s} \approx 3.77 \, \mathrm{m}\]
04

Review and Conclusion

The period of the oscillation is approximately \(0.0111 \, \mathrm{s}\) and the wavelength of the sound is approximately \(3.77 \, \mathrm{m}\).

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Key Concepts

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

Frequency
Frequency is a fundamental concept in understanding sound waves. It represents the number of complete oscillations a wave makes in a second. Frequency is measured in hertz (Hz), where one hertz corresponds to one cycle per second.
In the context of the exercise, a frequency of 90 Hz means the sand dunes' oscillation cycle repeats 90 times every second. This is a relatively low frequency, often perceived as a booming or rumbling sound.
When thinking about frequency:
  • High frequency means the sound wave oscillates more times per second, producing a higher-pitched sound.
  • Low frequency results in fewer oscillations per second, leading to a deeper sound, like the rumbling from the sand dunes.
Understanding frequency helps us unravel other characteristics of sound, like its period, which is the duration of one complete cycle.
Wavelength
Wavelength is a crucial element in the study of waves, including sound waves. It is the distance between two consecutive points in phase on a wave, such as crest-to-crest or trough-to-trough, typically measured in meters.
Based on the step-by-step solution given, the wavelength can be calculated from the formula \( \lambda = v \cdot T \), where \( v \) is the speed of sound, and \( T \) is the period of oscillation. Here, the wavelength of the sand dune's sound was found to be approximately 3.77 meters.
Consider these points about wavelength:
  • A longer wavelength corresponds to a lower frequency sound, which is deeper and travels farther.
  • A shorter wavelength corresponds to a higher frequency, often resulting in a high-pitched sound that does not travel as far.
Understanding wavelength is key to comprehending the sound's behavior as it travels through various media, such as air or even the sand itself.
Speed of Sound
The speed of sound is a measure of how quickly sound waves travel through a medium, such as air, water, or solids. It varies based on environmental conditions and the nature of the medium.
In the exercise, the speed of sound was assumed to be 340 m/s, a typical speed in dry air at 20°C. This was used to calculate the wavelength of the sound produced by the sand dunes.
Key factors about the speed of sound:
  • It increases with temperature since warmer air makes molecules move faster, thus transmitting sound more quickly.
  • Sound travels faster in solids and liquids than in gases because particles are closer together, facilitating quicker energy transfer.
The speed of sound is essential for calculations like determining wavelengths and understanding how sound travels over distances, such as the 10 km reach mentioned for the booming sound of sand dunes.

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Most popular questions from this chapter

A person on a railroad car blows a trumpet note at \(440 \mathrm{~Hz}\). The car is moving toward a wall at \(20.0 \mathrm{~m} / \mathrm{s}\). Find the sound frequency (a) at the wall and (b) reflected back to the trumpeter.

On July \(10,1996,\) a granite block broke away from a wall in Yosemite Valley and, as it began to slide down the wall, was launched into projectile motion. Seismic waves produced by its impact with the ground triggered seismographs as far away as \(200 \mathrm{~km} .\) Later measurements indicated that the block had a mass between \(7.3 \times 10^{7} \mathrm{~kg}\) and \(1.7 \times 10^{8} \mathrm{~kg}\) and that it landed \(500 \mathrm{~m}\) vertically below the launch point and \(30 \mathrm{~m}\) horizontally from it. (The launch angle is not known.) (a) Estimate the block's kinetic energy just before it landed. Consider two types of seismic waves that spread from the impact point - a hemispherical body wave traveled through the ground in an expanding hemisphere and a cylindrical surface wave traveled along the ground in an expanding shallow vertical cylinder (Fig. 17-49). Assume that the impact lasted \(0.50 \mathrm{~s}\), the vertical cylinder had a depth \(d\) of \(5.0 \mathrm{~m},\) and each wave type received \(20 \%\) of the energy the block had just before impact. Neglecting any mechanical energy loss the waves experienced as they traveled, determine the intensities of (b) the body wave and (c) the surface wave when they reached a seismograph \(200 \mathrm{~km}\) away. (d) On the basis of these results, which wave is more easily detected on a distant seismograph?

Two sounds differ in sound level by \(1.00 \mathrm{~dB}\). What is the ratio of the greater intensity to the smaller intensity?

A stationary motion detector sends sound waves of frequency \(0.150 \mathrm{MHz}\) toward a truck approaching at a speed of \(45.0 \mathrm{~m} / \mathrm{s}\). What is the frequency of the waves reflected back to the detector?

A police car is chasing a speeding Porsche 911 . Assume that the Porsche's maximum speed is \(80.0 \mathrm{~m} / \mathrm{s}\) and the police car's is \(54.0 \mathrm{~m} / \mathrm{s}\). At the moment both cars reach their maximum speed, what frequency will the Porsche driver hear if the frequency of the police car's siren is \(440 \mathrm{~Hz}\) ? Take the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\).

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