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Chapter 16: Problem 31

For research purposes a sonic buoy is tethered to the ocean floor and emits an infrasonic pulse of sound (speed \(=1522 \mathrm{m} / \mathrm{s}\) ). The period of this sound is 71 ms. Determine the wavelength of the sound.

Short Answer

Expert verified
The wavelength of the sound is 108.96 meters.

Step by step solution

01

Convert the Period to Seconds

The period of the sound is given as 71 milliseconds (ms). To work in the International System of Units, we first convert this into seconds (s). Thus, 71 ms is \(71 \text{ ms} = 71 \times 10^{-3} \text{ s} = 0.071 \text{ s}\).
02

Use the Wave Speed Formula

The speed of sound is given as \(v = 1522 \text{ m/s}\). Using the formula \(\lambda = v \times T \), where \(\lambda\) is the wavelength and \(T\) is the period, we can find the wavelength.
03

Calculate the Wavelength

Substitute the values for the speed of sound and the period into the formula:\(\lambda = 1522 \text{ m/s} \times 0.071 \text{ s}\).Calculating this gives \(\lambda = 108.962 \text{ m}\).
04

Round the Wavelength

Finally, round the wavelength to two decimal places for simplicity, giving us \(\lambda = 108.96 \text{ m}\).

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Key Concepts

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

Understanding Wave Speed
Wave speed is a crucial concept in physics, representing how fast a wave travels through a medium. It is commonly expressed in meters per second (m/s). For sound waves, the speed can vary depending on several factors like the medium through which it travels and the temperature of that medium. **Key Points about Wave Speed:**
  • The formula to calculate wave speed \[ v = \frac{\lambda}{T} \]where \( v \) is the wave speed, \( \lambda \) is the wavelength, and \( T \) is the period of the wave.
  • Sound travels faster through solids compared to liquids and gases because particles in solids are closer together.
  • For sound waves in the ocean, including the given sonic buoy scenario, the speed is often standardized like the provided speed of 1522 m/s.
The wave speed can be affected by environmental conditions such as pressure and temperature, which can cause the speed to vary slightly. Understanding wave speed is essential because it allows us to determine the other wave characteristics, such as wavelength, when the period is known. This connection is vital for applications such as sonar and marine exploration.
Characteristics of Sound Waves
Sound waves are a type of mechanical wave that requires a medium, like air, water, or solids, to travel. Unlike electromagnetic waves, which can travel in a vacuum, sound waves need these media to propagate. **Main Characteristics of Sound Waves:**
  • **Longitudinal Waves:** Sound waves cause the particles in the medium to vibrate back and forth in the direction the wave is traveling.
  • **Frequency and Period:** The frequency of a sound wave determines its pitch, while the period is the time it takes for one complete cycle of the wave.
  • **Amplitude:** This determines the loudness of the sound. Greater amplitude means a louder sound.
Infrasonic sounds, like the one emitted by the sonic buoy, have frequencies below the normal human hearing range. Such sounds can travel long distances through water, making them useful for research purposes. This characteristic makes sound waves invaluable tools for marine exploration and the study of underwater ecosystems.
Exploring the Period of a Wave
The period of a wave is the time it takes for one complete cycle of the wave to pass a given point. It's typically measured in seconds. The period is inversely related to the frequency, which measures how many cycles occur in one second.**Fundamental Aspects of Wave Period:**
  • **Formula:** The period \( T \) of a wave is calculated by \[ T = \frac{1}{f} \]where \( f \) is the frequency in hertz (Hz).
  • **Wave Equation:** In the exercise, we see a direct application of the period in calculating wavelength using \[ \lambda = v \times T \].
  • **Conversion:** Time units for period, like milliseconds, often need converting to seconds for standard calculations, as seen from the conversion of 71 ms to 0.071 s.
Understanding the period is key to solving many wave-related problems, as it helps in determining the wavelength when wave speed is known. For students and researchers, mastering the period concept provides a solid foundation for tackling complex wave phenomena in physics.

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Most popular questions from this chapter

A typical adult ear has a surface area of \(2.1 \times 10^{-3} \mathrm{m}^{2}\). The sound intensity during a normal conversation is about \(3.2 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) at the listener's ear. Assume that the sound strikes the surface of the ear perpendicularly. How much power is intercepted by the ear?

The tension in a string is \(15 \mathrm{N}\), and its linear density is \(0.85 \mathrm{kg} / \mathrm{m}\). A wave on the string travels toward the \(-x\) direction; it has an amplitude of \(3.6 \mathrm{cm}\) and a frequency of \(12 \mathrm{Hz}\). What are the (a) speed and (b) wavelength of the wave? (c) Write down a mathematical expression (like Equation 16.3 or 16.4 ) for the wave, substituting numbers for the variables \(A, f,\) and \(\lambda\)

One of the most important concepts we encountered in this chapter is the transverse wave. For instance, transverse waves travel along a guitar string when it is plucked or along a violin string when it is bowed. Problem 112 reviews how the travel speed depends on the properties of the string and on the tension in it. Problem 113 illustrates how the Doppler effect arises when an observer is moving away from or toward a stationary source of sound. In fact, we will see that it's possible for both situations to occur at the same time. A siren, mounted on a tower, emits a sound whose frequency is \(2140 \mathrm{Hz}\). A person is driving a car away from the tower at a speed of \(27.0 \mathrm{m} / \mathrm{s}\). As the figure illustrates, the sound reaches the person by two paths: the sound reflected from the building in front of the car, and the sound coming directly from the siren. The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). Concepts: (i) One way that the Doppler effect can arise is that the wavelength of the sound changes. For either the direct or reflected sound, does the wavelength change? (ii) Why does the driver hear a frequency for the reflected sound that is different from \(2140 \mathrm{Hz},\) and is it greater or smaller than \(2140 \mathrm{Hz} ?\) (iii) Why does the driver hear a frequency for the direct sound that is different from \(2140 \mathrm{Hz},\) and is it greater or smaller than \(2140 \mathrm{Hz}\) ? Calculations: What frequency does the person hear for the (a) reflected and (b) direct sound?

The siren on an ambulance is emitting a sound whose frequency is \(2450 \mathrm{Hz}\). The speed of sound is \(343 \mathrm{m} / \mathrm{s}\). (a) If the ambulance is stationary and you (the "observer") are sitting in a parked car, what are the wavelength and the frequency of the sound you hear? (b) Suppose that the ambulance is moving toward you at a speed of \(26.8 \mathrm{m} / \mathrm{s}\). Determine the wavelength and the frequency of the sound you hear. (c) If the ambulance is moving toward you at a speed of \(26.8 \mathrm{m} / \mathrm{s}\) and you are moving toward it at a speed of \(14.0 \mathrm{m} / \mathrm{s}\), find the wavelength and frequency of the sound you hear.

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at a location 22 \(\mathrm{m}\) away from the sound source is \(3.0 \times 10^{-4} \mathrm{W} / \mathrm{m}^{2} .\) What is the intensity at a spot that is \(78 \mathrm{m}\) away?
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