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The speed of light waves in air is greater than the speed of sound in air by about a factor of a million. Given a sound wave and a light wave of the same wavelength, both traveling through air, which statement about their frequencies is true? a) The frequency of the sound wave will be about a million times greater than that of the light wave. b) The frequency of the sound wave will be about a thousand times greater than that of the light wave. c) The frequency of the light wave will be about a thousand times greater than that of the sound wave. d) The frequency of the light wave will be about a million times greater than that of the sound wave. e) There is insufficient information to determine the relationship between the two frequencies.

Short Answer

Expert verified
a) The frequency of the light wave will be about the same as that of the sound wave. b) The frequency of the light wave will be about a hundred times greater than that of the sound wave. c) The frequency of the light wave will be about a thousand times greater than that of the sound wave. d) The frequency of the light wave will be about a million times greater than that of the sound wave. Answer: d) The frequency of the light wave will be about a million times greater than that of the sound wave.

Step by step solution

01

Recall the wave speed formula.

The wave speed formula relates the speed(v), frequency(f), and wavelength(λ) of a wave: v = fλ
02

Write the given information

We are given that the speed of light waves in air is greater than the speed of sound in air by a factor of a million, meaning: Speed of light (v_l) = 1,000,000 × Speed of sound (v_s) And, we are also given that both light and sound waves have the same wavelength: λ_l = λ_s
03

Use the wave speed formula for both waves

Use the wave speed formula to express the frequency of light waves (f_l) and sound waves (f_s) in terms of their speeds and wavelengths: For light waves: v_l = f_l λ_l For sound waves: v_s = f_s λ_s
04

Relate the two equations and solve for the frequency ratio of light and sound waves

As λ_l = λ_s, we can divide the light wave equation by the sound wave equation: ( v_l = f_l λ_l ) / ( v_s = f_s λ_s ) Which simplifies to: (v_l / v_s) = (f_l / f_s) We know that (v_l / v_s) = 1,000,000, so: 1,000,000 = (f_l / f_s)
05

Interpret the result and find the correct answer.

This result shows that the frequency of the light wave is a million times greater than the frequency of the sound wave. Therefore, the correct answer is: d) The frequency of the light wave will be about a million times greater than that of the sound wave.

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Key Concepts

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

Frequency of Waves
The frequency of a wave is a fundamental characteristic that describes how often the wave oscillates per unit of time. It is measured in Hertz (Hz), which equates to the number of wave cycles occurring per second. When considering waves traveling through a medium, the formula \( v = f\lambda \) is key. It connects three parameters: speed (v), frequency (f), and wavelength (\( \lambda \)). The frequency remains constant when the wave changes media, but the speed and wavelength adjust correspondingly.
This change directly impacts the perceived sound or light, as frequency determines the pitch of sound and the color of light. For example, higher frequencies in sound produce higher pitches, while in light they yield colors towards the violet end of the spectrum. Understanding frequency helps predict how waves behave across different environments.
Speed of Light
The speed of light in a vacuum is a universal constant, approximately \( 299,792,458 \) meters per second. However, when light travels through different media, such as air or glass, its speed reduces slightly. In air, light travels nearly at its maximum speed due to minimal resistance and refraction. This speed is vast and countless times greater than that of most other waves like sound.
In practical applications, the speed of light serves as a benchmark for developing technologies like GPS, telecommunications, and high-speed Internet. It's also used in calculations involving astronomical distances, reflecting the wonders and scales of our universe. Light's speed is why we perceive events almost instantaneously over distances on Earth, providing our understanding of the "speed of information."
Speed of Sound
Sound travels through air by creating pressure waves that move molecules. The speed of sound is around 343 meters per second at room temperature; however, this speed can vary depending on the medium. Sound travels faster in denser media, like water or steel, due to better particle connectivity allowing quicker vibration transfer.
The speed of sound is crucial in many areas, such as acoustics, sonar, and engineering. It influences how music is composed for different environments or how engineers design auditorium acoustics to ensure high-quality sound delivery. In the atmosphere, sound's speed affects phenomena like sonic booms produced when objects break the sound barrier.
  • Denser media: faster sound travel
  • Temperature and humidity affect speed
Understanding the speed of sound aids in applications ranging from music production to aviation.
Wavelength
Wavelength is a spatial characteristic of waves, representing the distance between successive crests or identical points in a wave. It is denoted by the Greek letter lambda (\( \lambda \)) and usually measured in meters. Wavelength is inversely related to frequency, meaning as wavelength increases, frequency decreases, given a constant wave speed. This relationship is essential when applying the formula \( v = f\lambda \).
Wavelength determines numerous properties of waves. In light, it defines color; in sound, it relates to pitch. Longer wavelengths in sound are associated with lower pitches and bass tones, while shorter wavelengths correspond to higher pitches. In light, short wavelengths appear as violets or blues, whereas longer wavelengths appear red.
Understanding wavelength helps in sectors like astronomy, where it aids in identifying light from distant galaxies, or telecommunications, designing networks based on signal clarity and distance.

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Most popular questions from this chapter

The middle-C key (key 52 ) on a piano corresponds to a fundamental frequency of about \(262 \mathrm{~Hz},\) and the sopranoC key (key 64) corresponds to a fundamental frequency of \(1046.5 \mathrm{~Hz}\). If the strings used for both keys are identical in density and length, determine the ratio of the tensions in the two strings.

A traveling wave propagating on a string is described by the following equation: $$ y(x, t)=(5.00 \mathrm{~mm}) \sin \left(\left(157.08 \mathrm{~m}^{-1}\right) x-\left(314.16 \mathrm{~s}^{-1}\right) t+0.7854\right) $$ a) Determine the minimum separation, \(\Delta x_{\min }\), between two points on the string that oscillate in perfect opposition of phases (move in opposite directions at all times). b) Determine the separation, \(\Delta x_{A B}\), between two points \(A\) and \(B\) on the string, if point \(B\) oscillates with a phase difference of 0.7854 rad compared to point \(A\). c) Find the number of crests of the wave that pass through point \(A\) in a time interval \(\Delta t=10.0 \mathrm{~s}\) and the number of troughs that pass through point \(B\) in the same interval. d) At what point along its trajectory should a linear driver connected to one end of the string at \(x=0\) start its oscillation to generate this sinusoidal traveling wave on the string?

The \(1 / R^{2}\) dependency for intensity can be thought of to be due to the fact that the same power is being spread out over the surface of a larger and larger sphere. What happens to the intensity of a sound wave inside an enclosed space, say a long hallway?

A string with linear mass density \(\mu=0.0250 \mathrm{~kg} / \mathrm{m}\) under a tension of \(T=250 . \mathrm{N}\) is oriented in the \(x\) -direction. Two transverse waves of equal amplitude and with a phase angle of zero (at \(t=0\) ) but with different frequencies \((\omega=3000\). rad/s and \(\omega / 3=1000 . \mathrm{rad} / \mathrm{s}\) ) are created in the string by an oscillator located at \(x=0 .\) The resulting waves, which travel in the positive \(x\) -direction, are reflected at a distant point, so there is a similar pair of waves traveling in the negative \(x\) -direction. Find the values of \(x\) at which the first two nodes in the standing wave are produced by these four waves.

You and a friend are holding the two ends of a Slinky stretched out between you. How would you move your end of the Slinky to create (a) transverse waves or (b) longitudinal waves?

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