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Chapter 19: Problem 28

A sound wave in a fluid medium is reflected at a barrier so that a standing wave is formed. The distance between nodes is \(3.84 \mathrm{~cm}\) and the speed of propagation is \(1520 \mathrm{~m} / \mathrm{s}\). Find the frequency.

Short Answer

Expert verified
The frequency of the standing wave in the fluid is approximately \(19805 Hz\).

Step by step solution

01

Convert the Distance between Nodes into Wavelength

The distance between nodes in the standing wave is given as \(3.84 cm\). However, this represents half a wavelength. So, the total wavelength \(\lambda\) is \(2 \times 3.84 cm = 7.68 cm\). Convert it to meters by dividing by 100, so \(\lambda = 0.0768~m\).
02

Calculate the Frequency

The speed of sound \(v\) in the medium is \(1520 m/s\), and it equals the product of the wavelength \(\lambda\) and the frequency \(f\), i.e., \(v = f \cdot \lambda\). Substituting the given values we can find the frequency \(f\) by re-arranging to \(f = v / \lambda\). This gives \(f = 1520~m/s / 0.0768~m = 19805~Hz\).

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Key Concepts

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

Understanding Standing Waves
Standing waves are a unique type of wave that occurs when two waves of the same frequency and amplitude travel in opposite directions through a medium and interfere with each other. These waves are characterized by their stationary pattern, consisting of nodes and antinodes.

- **Nodes:** Points along the medium where no movement occurs. They are formed because the interference between the two waves results in destructive interference at these locations.

- **Antinodes:** Points where the maximum displacement occurs. These are located halfway between nodes and result from constructive interference.

In the context given, the distance between nodes is crucial for calculating the wavelength. Remember, this distance represents half of the wavelength. By recognizing this fact, you can find the total wavelength essential for other calculations.
Measuring the Speed of Sound
The speed of sound in a particular medium determines how fast sound waves travel through it. In general, sound speed depends on several factors, including:

- **Medium type:** Sound travels faster through solids, then liquids, and slowest through gases.

- **Temperature:** Warmer temperatures generally increase sound speed because the molecules move more quickly.

Knowing that the speed of sound in our problem is given as \(1520~m/s\), this implies we're likely dealing with a liquid medium. In practice, using the speed of sound is vital for calculating other properties, such as wavelength and frequency of the standing wave.
Wavelength Calculation and Its Role
Wavelength is a fundamental measurement in the study of waves, defined as the distance between two consecutive points in phase on a wave, such as crest to crest or trough to trough.

In the context of standing waves in the exercise, the wavelength \(\lambda\) plays a crucial role in linking speed and frequency through the wave equation: \(v = f \cdot \lambda\). By knowing any two of these properties, you can compute the third.

For example, in the exercise provided, the halved distance between nodes is initially \(3.84~cm\). Converting this to full wavelength involves doubling this value, resulting in \(7.68~cm\) or \(0.0768~m\) in meters. This wavelength then connects with the wave speed to find frequency: \(f = v / \lambda = 1520~m/s / 0.0768~m\), leading to a calculated frequency of \(19805~Hz\).

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Most popular questions from this chapter

A continuous sinusoidal longitudinal wave is sent along a coiled spring from a vibrating source attached to it. The frequency of the source is \(25 \mathrm{~Hz}\), and the distance between successive rarefactions in the spring is \(24 \mathrm{~cm} .(a)\) Find the wave speed. (b) If the maximum longitudinal displacement of a particle in the spring is \(0.30 \mathrm{~cm}\) and the wave moves in the \(-x\) direction, write the equation for the wave. Let the source be at \(x=0\) and the displacement \(s=0\) at the source when \(t=0\).

Find the intensity ratio of two sounds whose sound levels differ by \(1.00 \mathrm{~dB}\).

A bat is flitting about in a cave, navigating very effectively by the use of ultrasonic bleeps (short emissions of highfrequency sound lasting a millisecond or less and repeated several times a second). Assume that the sound emission frequency of the bat is \(39.2 \mathrm{kHz}\). During one fast swoop directly toward a flat wall surface, the bat is moving at \(8.58 \mathrm{~m} / \mathrm{s}\). Calculate the frequency of the sound the bat hears reflected off the wall.

A siren emitting a sound of frequency \(1000 \mathrm{~Hz}\) moves away from you toward a cliff at a speed of \(10.0 \mathrm{~m} / \mathrm{s} .(a)\) What is the frequency of the sound you hear coming directly from the siren? ( \(b\) ) What is the frequency of the sound you hear reflected off the cliff? ( \(c\) ) Find the beat frequency. Could you hear the beats? Take the speed of sound in air as \(330 \mathrm{~m} / \mathrm{s}\)

A person in a car blows a trumpet sounding at \(438 \mathrm{~Hz}\). The car is moving toward a wall at \(19.3 \mathrm{~m} / \mathrm{s} .\) Calculate \((a)\) the frequency of the sound as received at the wall and \((b)\) the frequency of the reflected sound arriving back at the source.
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