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A sinusoidal wave of frequency \(500 \mathrm{~Hz}\) has a speed of \(350 \mathrm{~m} / \mathrm{s}\). (a) How far apart are two points that differ in phase by \(\pi / 3\) rad? (b) What is the phase difference between two displacements at a certain point at times \(1.00 \mathrm{~ms}\) apart?

Short Answer

Expert verified
(a) 0.1166 m; (b) \( \pi \) rad.

Step by step solution

01

Determine the Wavelength

First, find the wavelength \( \lambda \) using the relationship \( v = f \lambda \), where \( v = 350 \, \mathrm{m/s} \) is the wave speed and \( f = 500 \, \mathrm{Hz} \) is the frequency.\[ \lambda = \frac{v}{f} = \frac{350}{500} = 0.7 \, \mathrm{m} \]
02

Calculate Distance for Phase Difference

Now, calculate the distance \( d \) corresponding to a phase difference \( \Delta \phi = \frac{\pi}{3} \) rad. The phase difference \( \Delta \phi \) and the distance \( d \) are related by \( \Delta \phi = \frac{2\pi}{\lambda} d \). Solving for \( d \):\[ d = \frac{\Delta \phi \cdot \lambda}{2\pi} = \frac{(\pi/3) \cdot 0.7}{2\pi} \approx 0.1166 \, \mathrm{m} \]
03

Calculate Angular Frequency

Find the angular frequency \( \omega \) using \( \omega = 2\pi f \).\[ \omega = 2\pi \times 500 = 1000\pi \, \mathrm{rad/s} \]
04

Determine Phase Difference in Time

Calculate the phase difference \( \Delta \phi_t \) for a time difference \( \Delta t = 1.00 \, \mathrm{ms} = 0.001 \, \mathrm{s} \) using \( \Delta \phi_t = \omega \Delta t \).\[ \Delta \phi_t = 1000\pi \times 0.001 = \pi \, \mathrm{rad} \]

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Key Concepts

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

Frequency
Frequency is a fundamental concept in wave physics, describing how often a wave oscillates in a given period of time. In simple terms, the frequency of a wave is the number of complete cycles that occur per second. It's measured in Hertz (Hz), where 1 Hz corresponds to one cycle per second.
For example, in our problem, the frequency of the sinusoidal wave is given as 500 Hz. This means that 500 cycles of the wave pass through a point every second. Frequency is crucial because it helps us determine other properties of the wave, such as wavelength and speed.
  • Higher frequency means more cycles per second.
  • Frequency is inversely proportional to wavelength, given a constant wave speed.
  • It's essential in applications like audio, radios, and even in measuring seismic activities.
Understanding frequency helps in identifying and distinguishing different types of waves and their behaviors.
Wavelength
Wavelength, often denoted by the symbol \( \lambda \), is the distance between two successive identical points of a wave, such as crest to crest or trough to trough. It is a key spatial characteristic of a wave and can be thought of as the length of one complete cycle in the wave.
In our given exercise, the wavelength is calculated using the formula \( \lambda = \frac{v}{f} \), where \( v \) is the wave speed and \( f \) is the frequency. For a wave speed of 350 m/s and a frequency of 500 Hz, the wavelength is 0.7 meters.
  • Wavelength is inversely proportional to frequency for a constant wave speed: as frequency increases, wavelength decreases.
  • Different wavelengths are perceived differently by our senses, for example, different colors in light waves and different pitches in sound waves.
  • Knowing the wavelength helps in calculating the wave's energy and phase difference for various points along the wave.
Wavelength is applicable in diverse fields such as telecommunications, optics, and even quantum mechanics, making it a versatile and necessary concept.
Phase Difference
Phase difference refers to the difference in phase angle between two points along a wave or between two waves. It tells us how "out of step" two waves are with each other. Phase can be thought of as the stage in the cycle that the wave is at: beginning, middle, or end. It's measured in radians or degrees.
In the provided problem, you are asked to find the distance corresponding to a phase difference of \( \pi/3 \) rad. This indicates that one point on the wave is at a different stage in its cycle compared to another point. The formula \( \Delta \phi = \frac{2\pi}{\lambda} d \) is used to relate phase difference to physical distance.
  • A full phase difference of \( 2\pi \) radians corresponds to one full cycle, meaning the points are in-phase after each cycle.
  • Phase differences are used in interference and diffraction studies in waves to predict patterns and behaviors.
  • In applications, phase difference is crucial for the functioning of devices like oscilloscopes and in understanding wave behaviors in mediums.
Phase difference is a subtle concept but powerful in understanding wave interactions and predicting outcomes in wave-related phenomena.
Angular Frequency
Angular frequency, symbolized as \( \omega \), gives us a measure of how many radians a wave covers in one second. Unlike standard frequency, which measures cycles per second, angular frequency measures the angle in radians that a wave travels through in a second, and is closely related to the frequency by the formula \( \omega = 2\pi f \).
For the wave in the exercise, the angular frequency is computed as \( 1000\pi \) rad/s.
  • Angular frequency helps in understanding the oscillatory motion more deeply, especially in circular motion contexts.
  • It is pivotal in physics for solving differential equations related to harmonic motion.
  • Many fields, like engineering and acoustics, use angular frequency to describe phenomena that involve oscillations or rotations.
This concept is most useful when we deal with problems that involve changes over time, such as calculating the phase difference over given time intervals. Angular frequency is a foundational element in both mechanics and electronics to understand the nuances of periodic motions.

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Most popular questions from this chapter

A sinusoidal transverse wave of amplitude \(y_{m}\) and wavelength \(\lambda\) travels on a stretched cord. (a) Find the ratio of the maximum particle speed (the speed with which a single particle in the cord moves transverse to the wave) to the wave speed. (b) Does this ratio depend on the material of which the cord is made?

Two sinusoidal waves of the same period, with amplitudes of 5.0 and \(7.0 \mathrm{~mm}\), travel in the same direction along a stretched string; they produce a resultant wave with an amplitude of \(9.0 \mathrm{~mm}\). The phase constant of the \(5.0 \mathrm{~mm}\) wave is \(0 .\) What is the phase constant of the \(7.0 \mathrm{~mm}\) wave?

A string oscillates according to the equation \(y^{\prime}=(0.50 \mathrm{~cm}) \sin \left[\left(\frac{\pi}{3} \mathrm{~cm}^{-1}\right) x\right] \cos \left[\left(40 \pi \mathrm{s}^{-1}\right) t\right]\) What are the (a) amplitude and (b) speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation? (c) What is the distance between nodes? (d) What is the transverse speed of a particle of the string at the position \(x=1.5 \mathrm{~cm}\) when \(t=\frac{9}{8} \mathrm{~s} ?\)

A string along which waves can travel is \(2.70 \mathrm{~m}\) long and has a mass of \(260 \mathrm{~g}\). The tension in the string is \(36.0 \mathrm{~N}\). What must be the frequency of traveling waves of amplitude \(7.70 \mathrm{~mm}\) for the average power to be \(85.0 \mathrm{~W} ?\)

Two sinusoidal waves with identical wavelengths and amplitudes travel in opposite directions along a string with a speed of \(10 \mathrm{~cm} / \mathrm{s}\). If the time interval between instants when the string is flat is \(0.50 \mathrm{~s}\), what is the wavelength of the waves?

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