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A sinusoidal sound wave moves at \(343 \mathrm{~m} / \mathrm{s}\) through air in the positive direction of an \(x\) axis. At one instant during the oscillations, air molecule \(A\) is at its maximum displacement in the negative direction of the axis while air molecule \(B\) is at its equilibrium position. The separation between those molecules is \(15.0 \mathrm{~cm}\), and the molecules between \(A\) and \(B\) have intermediate displacements in the negative direction of the axis. (a) What is the frequency of the sound wave? In a similar arrangement but for a different sinusoidal sound wave, at one instant air molecule \(C\) is at its maximum displacement in the positive direction while molecule \(D\) is at its maximum displacement in the negative direction. The separation between the molecules is again \(15.0 \mathrm{~cm}\), and the molecules between \(C\) and \(D\) have intermediate displacements. (b) What is the frequency of the sound wave?

Short Answer

Expert verified
(a) 571.67 Hz; (b) 1143.33 Hz

Step by step solution

01

Understand the Problem for Part (a)

The information given states that molecule A is at its maximum negative displacement and molecule B is at the equilibrium position. The separation between these molecules is 15.0 cm. The molecules between A and B have intermediate displacements in the negative direction, indicating that this distance represents a quarter of a wavelength.
02

Calculate the Wavelength for Part (a)

Since the separation of 15.0 cm represents a quarter of a wavelength (\(\frac{\lambda}{4}\)), we can set up the equation:\[\lambda = 4 \times 0.15 \text{ m} = 0.60 \text{ m}.\]
03

Determine Frequency for Part (a)

Using the speed of sound \(v = 343 \,\text{m/s}\) and the wavelength \(\lambda = 0.60 \,\text{m}\), use the wave equation \(v = f \cdot \lambda\) to find the frequency \(f:\)\[f = \frac{v}{\lambda} = \frac{343 \, \text{m/s}}{0.60 \, \text{m}} = 571.67 \, \text{Hz}.\]
04

Understand the Problem for Part (b)

In this scenario, molecule C is at its maximum positive displacement and molecule D at its maximum negative displacement with a separation of 15.0 cm. This separation between C and D represents half of a wavelength because they are at opposite maximum displacements.
05

Calculate Wavelength for Part (b)

Since 15.0 cm is half a wavelength (\(\frac{\lambda}{2}\)), we can find the wavelength \(\lambda\) as follows:\[\lambda = 2 \times 0.15 \text{ m} = 0.30 \text{ m}.\]
06

Determine Frequency for Part (b)

Using the wave equation with the new wavelength:\[f = \frac{v}{\lambda} = \frac{343 \, \text{m/s}}{0.30 \, \text{m}} = 1143.33 \, \text{Hz}.\]

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Key Concepts

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

Frequency
Frequency is the number of oscillations or vibrations that occur per unit time. In the context of sound waves, frequency is a measure of how many complete cycles of the wave pass a certain point every second. Frequency is measured in Hertz (Hz), which corresponds to cycles per second. A higher frequency means that the sound wave oscillates more quickly.

When calculating the frequency of a sound wave, you use the wave equation:
  • Velocity of the wave (\(v\))
  • Wavelength (\(\lambda\))
  • Frequency (\(f\))
The formula is expressed as:\[f = \frac{v}{\lambda}\]This formula shows that frequency is directly related to both the speed of the wave and inversely related to the wavelength of the wave. Essentially, if you know how fast a wave travels (its velocity) and the wavelength, you can determine its frequency.

In the presented exercise, understanding whether the separation between molecules represents a quarter or half a wavelength is crucial to calculate the frequency accurately. For example, in part (a) of the exercise, a separation of 15.0 cm is one-quarter the wavelength, leading us to calculate a frequency of 571.67 Hz. In contrast, a half-wavelength separation in part (b) results in a different frequency of 1143.33 Hz.
Wavelength
Wavelength is the distance between successive crests (peaks) or troughs (valleys) of a wave. In simpler terms, it's how long a single full wave cycle is. Wavelength is often denoted by the Greek letter lambda (\(\lambda\)) and measured in meters (m).

In the context of sound waves, wavelength is significant because it influences the sound's frequency and how it moves through a medium. Longer wavelengths mean lower frequencies and thus lower-pitched sounds, while shorter wavelengths correspond to higher frequencies and higher pitches.

In practical terms, when working with sound waves, the relationship between wavelength and the position of molecules gives insight into the wave's behavior. For instance, when one molecule is at its maximum position while another is at equilibrium, the distance between them informs the part of the wavelength represented.

In the exercise, Part A used a quarter-wavelength concept, while Part B required understanding of what half a wavelength means throughout the wave pattern. Calculating these correctly allows you to determine how they correspond to changes in frequency and sound characteristics.
Wave Equation
The wave equation is a fundamental principle used to relate the speed, frequency, and wavelength of a wave. It is an essential tool in understanding how different properties of waves interact. The equation is:\[v = f \cdot \lambda\]Where:
  • \(v\): Velocity or speed of the wave in meters per second (m/s).
  • \(f\): Frequency in Hertz (Hz).
  • \(\lambda\): Wavelength in meters (m).
Using this formula, if you know any two properties, you can solve for the third. For example, knowing the speed and wavelength allows you to find the frequency. In sound wave problems, the speed is often known, and finding either frequency or wavelength is the goal.

In scenarios like the provided exercise, interpreting the wave's behavior and its effects on molecule displacements provides insight into the wave equation's application. Calculating wavelength accurately based on displacement information helps effectively use the wave equation, ensuring precise determination of sound wave properties such as frequency.

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Most popular questions from this chapter

Pipe \(A\), which is \(1.20 \mathrm{~m}\) long and open at both ends, oscillates at its third lowest harmonic frequency. It is filled with air for which the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). Pipe \(B\), which is closed at one end, oscillates at its second lowest harmonic frequency. This frequency of \(B\) happens to match the frequency of \(A .\) An \(x\) axis extends along the interior of \(B\), with \(x=0\) at the closed end. (a) How many nodes are along that axis? What are the (b) smallest and (c) second smallest value of \(x\) locating those nodes? (d) What is the fundamental frequency of \(B\) ?

Ultrasound, which consists ultrasour of sound waves with frequencies above the human audible range, can Artery be used to produce an image of the interior of a human body. Moreover, ultrasound can be used to measure Figure 17-47 Problem \(83 .\) the speed of the blood in the body; it does so by comparing the frequency of the ultrasound sent into the body with the frequency of the ultrasound reflected back to the body's surface by the blood. As the blood pulses, this detected frequency varies. Suppose that an ultrasound image of the arm of a patient shows an artery that is angled at \(\theta=20^{\circ}\) to the ultrasound's line of travel (Fig. 17-47). Suppose also that the frequency of the ultrasound reflected by the blood in the artery is increased by a maximum of \(5495 \mathrm{~Hz}\) from the original ultrasound frequency of \(5.000000 \mathrm{MHz} .\) (a) In Fig. \(17-47\), is the direction of the blood flow rightward of leftward? (b) The speed of sound in the human arm is \(1540 \mathrm{~m} / \mathrm{s}\). What is the maximum speed of the blood? (Hint: The Doppler effect is caused by the component of the blood's velocity along the ultrasound's direction of travel.) (c) If angle \(\theta\) were greater, would the reflected frequency be greater or less?

A man strikes one end of a thin rod with a hammer. The speed of sound in the rod is 15 times the speed of sound in air. A woman, at the other end with her ear close to the rod, hears the sound of the blow twice with a \(0.12\) s interval between; one sound comes through the rod and the other comes through the air alongside the rod. If the speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\), what is the length of the rod?

Hot chocolate effect. Tap a metal spoon inside a mug of water and note the frequency \(f_{i}\) you hear. Then add a spoonful of powder (say, chocolate mix or instant coffee) and tap again as you stir the powder. The frequency you hear has a lower value \(f_{s}\) because the tiny air bubbles released by the powder change the water's bulk modulus. As the bubbles reach the water surface and disappear, the frequency gradually shifts back to its initial value. During the effect, the bubbles don't appreciably change the water's density or volume or the sound's wavelength.

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.

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