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A sound wave in a fluid medium is reflected at a barrier so that a standing wave is formed. The distance between nodes is \(4.9 \mathrm{~cm}\), and the speed of propagation is \(1250 \mathrm{~m} / \mathrm{s}\). Find the frequency of the sound wave.

Short Answer

Expert verified
Frequency is approximately 12755.10 Hz.

Step by step solution

01

Understanding the Problem

We are given a standing wave in a fluid medium, where the distance between the nodes is known. The nodes are points of no displacement in a wave. We need to find the frequency of the wave given the speed of propagation.
02

Recall the Relationship between Wavelength and Node Distance

The distance between two consecutive nodes in a standing wave is half the wavelength, since nodes appear at positions that are multiples of half wavelengths. Thus, if the distance between nodes is given, that distance equals \( \frac{\lambda}{2} \) where \( \lambda \) is the wavelength.
03

Calculate the Wavelength

Given that the distance between nodes is \(4.9 \mathrm{~cm}\), or \(0.049 \mathrm{~m}\), we use the relationship \( \frac{\lambda}{2} = 0.049 \mathrm{~m} \) to find the wavelength \( \lambda \). So, \( \lambda = 2 \times 0.049 \mathrm{~m} = 0.098 \mathrm{~m} \).
04

Use the Wave Equation to Find the Frequency

The wave equation relates speed, frequency, and wavelength: \( v = f \lambda \). Rearranging to find frequency, we have \( f = \frac{v}{\lambda} \). Substituting the values \( v = 1250 \mathrm{~m/s} \) and \( \lambda = 0.098 \mathrm{~m} \), we get \( f = \frac{1250}{0.098} \approx 12755.10 \mathrm{~Hz} \).

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Key Concepts

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

Standing Waves
In our everyday lives, sound often travels in the form of standing waves. This phenomenon occurs when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other. The result is a wave pattern that appears to be "standing" still, hence the name "standing wave." Such waves can be found in musical instruments, where reflections create stable patterns.

A standing wave has specific points called nodes and antinodes.
  • Nodes are points that do not move; they remain stationary as the waves cancel each other out at these points.
  • Antinodes are points where the waves move the most, resulting in maximum displacement.
Interestingly, the distance between two consecutive nodes is exactly half of the wavelength of the wave. This property is crucial when analyzing and understanding how standing waves behave. Recognizing their formation and behavior helps us in various applications, such as determining characteristics like wavelength and frequency.
Wave Frequency
Wave frequency is a fundamental concept when studying sound waves. It tells us how many times a wave oscillates or completes a cycle in one second. This is measured in Hertz (Hz). For sound waves, frequency is directly related to the pitch of the sound—a higher frequency means a higher pitch.

Given a specific speed of propagation for a wave, we can calculate its frequency if we know its wavelength. The formula to remember is:- \[ f = \frac{v}{\lambda} \]where:
  • \( f \) is the frequency,
  • \( v \) is the speed of the wave, and
  • \( \lambda \) is the wavelength.
The right frequency calculation ensures that we understand the sound properties better, for example, in acoustics and audio engineering. Knowing the frequency helps us tailor sound experiences in various environments, such as concert halls and recording studios.
Wavelength Calculation
Understanding how to calculate the wavelength of a sound wave is essential in physics. In the context of standing waves, the nodes' distance is key. By definition, the distance between two nodes is half of the wavelength (\( \frac{\lambda}{2} \), where \( \lambda \) is the wavelength). Therefore, if we know the distance between nodes, finding the wavelength becomes straightforward.- Calculate the wavelength using:\[ \lambda = 2 imes ( ext{node distance}) \]In the given problem, the node distance was \(4.9 \text{ cm} \), or \(0.049 \text{ m} \), and thus:\[ \lambda = 2 imes 0.049 \text{ m} = 0.098 \text{ m} \]Knowing the wavelength allows us to connect other properties of the wave, such as frequency and speed. This relationship lays the groundwork for exploring more advanced concepts in wave dynamics and their applications. Understanding how wavelength plays into the larger wave equation helps engineers design better sound systems and more efficient communication technologies.

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Most popular questions from this chapter

From two sources, sound waves of frequency \(270 \mathrm{~Hz}\) are emitted in phase in the positive direction of an \(x\) axis. At a detector that is on the axis and \(5.00 \mathrm{~m}\) from one source and \(4.00 \mathrm{~m}\) from the other source, what is the phase difference between the waves (a) in radians and (b) as a multiple of the wavelength?

One of the harmonic frequencies of tube \(A\) with two open ends is \(400 \mathrm{~Hz}\). The next-highest harmonic frequency is \(480 \mathrm{~Hz}\). (a) What harmonic frequency is next highest after the harmonic frequency \(160 \mathrm{~Hz}\) ? (b) What is the number of this next-highest harmonic? One of the harmonic frequencies of tube \(B\) with only one open end is \(1080 \mathrm{~Hz}\). The next-highest harmonic frequency is \(1320 \mathrm{~Hz}\). (c) What harmonic frequency is next highest after the harmonic frequency \(600 \mathrm{~Hz}\) ? (d) What is the number of this nexthighest harmonic?

A \(2000 \mathrm{~Hz}\) siren and a civil defense official are both at rest with respect to the ground. What frequency does the official hear if the wind is blowing at \(15 \mathrm{~m} / \mathrm{s}\) (a) from source to official and (b) from official to source?

(a) Find the speed of waves on a violin string of mass \(860 \mathrm{mg}\) and length \(22.0 \mathrm{~cm}\) if the fundamental frequency is \(920 \mathrm{~Hz}\). (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string?

Organ pipe \(A\), with both ends open, has a fundamental frequency of \(425 \mathrm{~Hz}\). The fifth harmonic of organ pipe \(B\), with one end open, has the same frequency as the second harmonic of pipe \(A\). How long are (a) pipe \(A\) and (b) pipe \(B ?\)

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