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Chapter 16: Problem 9

A sound wave of frequency \(100 \mathrm{~Hz}\) is travelling in air. The speed of sound in air is \(350 \mathrm{~m} \mathrm{~s}^{-1}\). (a) By how much is the phase changed at a given point in \(2 \cdot 5 \mathrm{~ms} ?\) (b) What is the phase difference at a given instant between two points separated by a distance of \(10 \cdot 0 \mathrm{~cm}\) along the direction of propagation?

Short Answer

Expert verified
(a) Phase change in 2.5 ms is \( 0.5 \pi \). (b) Phase difference over 10 cm is \( 0.179 \pi \).

Step by step solution

01

Calculate Wavelength

To solve these problems, we begin by determining the wavelength \( \lambda \) of the wave, using the relation between speed \( v \), frequency \( f \), and wavelength: \[ v = f \times \lambda \]Given: - Speed \( v = 350 \, \text{m/s} \)- Frequency \( f = 100 \, \text{Hz} \)By rearranging the formula: \[ \lambda = \frac{v}{f} = \frac{350}{100} = 3.5 \, \text{m} \]
02

Calculate Phase Change Over Time

The phase change \( \Delta \phi \) at a point over a period of time \( t \) is given by:\[ \Delta \phi = 2\pi f t \]Given: - Frequency \( f = 100 \, \text{Hz} \)- Time \( t = 2.5 \times 10^{-3} \, \text{s} \)Substituting values:\[ \Delta \phi = 2\pi \times 100 \times 2.5 \times 10^{-3} = 2\pi \times 0.25 = 0.5 \pi \]Thus, the phase change at a given point in 2.5 ms is \( 0.5 \pi \) radians.
03

Calculate Phase Difference Over Distance

The phase difference \( \Delta \phi \) between points separated by distance \( d \) is given by:\[ \Delta \phi = \frac{2\pi}{\lambda} \times d \]Given:- Wavelength \( \lambda = 3.5 \, \text{m} \)- Distance \( d = 0.1 \, \text{m} \)Substituting values:\[ \Delta \phi = \frac{2\pi}{3.5} \times 0.1 = \frac{2\pi}{35} \approx 0.179 \pi \]The phase difference between two points separated by 10 cm is approximately \( 0.179 \pi \) radians.

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Key Concepts

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

Sound Waves
Sound waves are vibrations that travel through a medium, such as air, water, or solids. These vibrations are longitudinal waves, meaning that the vibrations occur in the same direction as the wave is traveling. Sound is created when an object causes particles in its surrounding medium to oscillate.
As the waves move, they carry energy and information, allowing us to hear music, speech, and other everyday noises.
  • Longitudinal Waves: Sound waves compress and expand the medium they travel through, making them different from transverse waves like light.
  • Medium Dependency: Sound requires a medium to propagate, which is why sound cannot travel in a vacuum.
Even though we often think of sound in terms of musical notes, sound waves also include a broad range of frequencies from very low bass to high-pitched sounds beyond human hearing capabilities.
Phase Change
Phase change in the context of sound waves refers to a shift in the wave's position relative to a reference point. This change is usually measured in degrees or radians. When sound waves travel, the phase can indicate how much of the wave cycle鈥攊ts peaks and troughs鈥攈as passed through the reference point.
A phase change can happen over time or through space:
  • Time-Based Phase Change: Relates to how the position of the wave shifts over a given period.
  • Distance-Based Phase Difference: Compares wave positions at two separate points in space.
Understanding phase change helps in analyzing wave superposition, interference patterns, and timing of sound events. It is especially useful in acoustics and audio engineering.
Wavelength Calculation
Wavelength (\( \lambda \)) is the distance over which the wave's shape repeats. For sound waves, the wavelength depends on the speed of sound in the medium and the frequency of the wave. By using the formula \( \lambda = \frac{v}{f} \), we can easily calculate the wavelength if the speed (\( v \)) and frequency (\( f \)) are known.
In our example:
  • Speed of Sound: Given as 350 m/s.
  • Frequency: Given as 100 Hz.
By substituting the values into the formula, the wavelength of the sound wave in air is calculated to be 3.5 meters. This shows how tightly or loosely packed the wave formations are in a medium.
Frequency
Frequency (\( f \)) indicates how many wave cycles occur in one second. It is measured in hertz (Hz). High frequencies mean more cycles per second, associated with higher-pitched sounds. Low frequencies have fewer cycles, producing lower-pitched sounds.
This example describes a frequency of 100 Hz, meaning each second contains 100 wave cycles. It's essential to understanding sound because:
  • Determines Pitch: Higher frequency leads to a higher pitch.
  • Affects Wavelength: Inverse relation to wavelength; as frequency increases, wavelength decreases.
Frequency not just describes sound but also plays a role in determining how sound interacts with other waves, making it crucial for technologies like audio compression and noise cancellation.
Speed of Sound
The speed of sound is the rate at which sound waves travel through a medium. In air, this speed is influenced by several factors, including temperature, pressure, and humidity. For the given problem, the speed of sound is provided as 350 m/s.
It's an essential concept because it allows us to calculate other properties of a wave, such as wavelength and phase differences over distances.
  • Medium Dependency: Different media enable sound to travel at different speeds (e.g., faster in water than in air).
  • Environmental Factors: Sound speed in air can change with temperature and pressure variations.
Knowing the speed of sound helps predict how sound behaves in different environments and is vital in fields such as meteorology and audio engineering.

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Most popular questions from this chapter

An open organ pipe has a length of \(5 \mathrm{~cm}\). (a) Find the fundamental frequency of vibration of this pipe. (b) What is the highest harmonic of such a tube that is in the audible range ? Speed of sound in air is \(340 \mathrm{~m} \mathrm{~s}^{-1}\) and the audible range is \(20-20,000 \mathrm{~Hz}\).

A copper rod of length \(1 \cdot 0 \mathrm{~m}\) is clamped at its middle point. Find the frequencies between \(20 \mathrm{~Hz}\) and \(20,000 \mathrm{~Hz}\) at which standing longitudinal waves can be set up in the rod. The speed of sound in copper is \(38 \mathrm{~km} \mathrm{~s}^{-1}\).

Three sources of sound \(S_{1}, S_{2}\) and \(S_{3}\) of equal intensity are placed in a straight line with \(S_{1} S_{2}=S_{2} S_{3}\) (figure 16-E5). At a point \(P\), far away from the sources, the wave coming from \(S_{2}\) is \(120^{\circ}\) ahead in phase of that from \(S_{1}\). Also, the wave coming from \(S_{3}\) is \(120^{\circ}\) ahead of that from \(S_{2}\). What would be the resultant intensity of sound at \(P ?\)

A small source of sound vibrating at frequency \(500 \mathrm{~Hz}\) is rotated in a circle of radius \(100 / \pi \mathrm{cm}\) at a constant angular speed of \(5 \cdot 0\) revolutions per second. A listener situates himself in the plane of the circle. Find the minimum and the maximum frequency of the sound observed. Speed of sound in air \(=332 \mathrm{~m} \mathrm{~s}^{-1}\).

Two electric trains run at the same speed of \(72 \mathrm{~km} / \mathrm{h}\) along the same track and in the same direction with a separation of \(2 \cdot 4 \mathrm{~km}\) between them. The two trains simultaneously sound brief whistles. A person is situated at a perpendicular distance of \(500 \mathrm{~m}\) from the track and is equidistant from the two trains at the instant of the whistling. If both the whistles were at \(500 \mathrm{~Hz}\) and the speed of sound in air is \(340 \mathrm{~m} / \mathrm{s}\), find the frequencies heard by the person.
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