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Chapter 17: Problem 9

Sinusoidal Wave Two 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 \(\pi / 3 \mathrm{rad} ?(\mathrm{~b})\) What is the phase difference between two displacements at a certain point at times \(1.00 \mathrm{~ms}\) apart?

Short Answer

Expert verified
a) 0.1167 m b) \( \pi \) rad

Step by step solution

01

- Identify Given Values

Frequency of the wave, \( f=500 \text{ Hz} \) Speed of the wave, \( v=350 \text{ m/s} \)
02

- Calculate Wavelength

Use the formula for the speed of a wave: \( v = f \lambda \) Rearrange to solve for the wavelength \( \lambda \): \( \lambda = \frac{v}{f} = \frac{350 \text{ m/s}}{500 \text{ Hz}} = 0.7 \text{ m} \)
03

- Calculate Distance for Phase Difference (Part a)

To find the distance corresponding to a phase difference of \( \frac{\pi}{3} \) radians, use the relationship between phase difference and distance: \( \Delta \phi = \frac{2\pi}{\lambda} \Delta x \) Solve for the distance \( \Delta x \): \( \Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} = \frac{(\frac{\pi}{3}) \cdot 0.7 \text{ m}}{2\pi} = \frac{0.7}{6} \text{ m} = 0.1167 \text{ m} \)
04

- Calculate Angular Frequency

The angular frequency \( \omega \) is calculated using: \( \omega = 2\pi f = 2\pi (500 \text{ Hz}) = 1000\pi \text{ rad/s} \)
05

- Calculate Phase Difference for Time Interval (Part b)

Use the formula for phase difference, which is dependent on time: \( \Delta \phi = \omega \Delta t \) Given \( \Delta t = 1.00 \text{ ms } = 1.00 \times 10^{-3} \text{ s} \): \( \Delta \phi = 1000\pi \cdot 1.00 \times 10^{-3} \text{ s} = \pi \text { rad} \)

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Key Concepts

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

Wave Frequency
Wave frequency is a fundamental concept when dealing with sinusoidal waves. Frequency, denoted by the symbol \( f \), refers to the number of cycles a wave completes in one second and is measured in Hertz (Hz). For example, in our exercise, the wave frequency is given as \(500 \text{ Hz}\). This means that every second, the wave completes 500 cycles.
Frequency relates directly to the pitch of sound waves; a higher frequency means a higher pitch, while a lower frequency means a lower pitch.
Key points to remember:
  • Frequency \( f \) determines how many wave cycles occur per second.
  • Measured in Hertz (Hz).
  • In sound waves, it affects pitch.

Frequency is essential for calculating other wave properties such as wavelength and angular frequency.
Wave Speed
Wave speed, denoted as \( v \), is the rate at which the wave propagates through a medium. It's measured in meters per second (m/s). In our exercise, the wave speed is \(350 \text{ m/s}\).
Wave speed can be determined using the formula \( v = f \lambda \) where \( f \) is the frequency and \( \lambda \) is the wavelength. This relationship shows how wave speed is influenced by both frequency and wavelength.
Key points to remember:
  • Wave speed \( v \) is how fast the wave travels through the medium.
  • Measured in meters per second (m/s).
  • Calculated using \( v = f \lambda \).
Wavelength Calculation
Wavelength, represented by \( \lambda \), is the distance between successive peaks or troughs of a wave. To find the wavelength, we use the formula: \( \lambda = \frac{v}{f} \). For our wave with a speed of \(350 \text{ m/s}\) and frequency \(500 \text{ Hz}\), the wavelength is calculated as:
\( \lambda = \frac{350 \text{ m/s}}{500 \text{ Hz}} = 0.7 \text{ m} \).
This means each cycle of the wave is 0.7 meters long.
Key points to remember:
  • Wavelength \( \lambda \) is the distance between identical points in adjacent cycles of a wave.
  • Measured in meters (m).
  • Calculated using \( \lambda = \frac{v}{f} \).
Phase Difference
Phase difference, \( \Delta \phi \), measures the shift between two points in terms of wave cycles. It is critical for understanding how much one part of the wave lags or leads another. In our exercise, we calculate the distance between two points that differ by a phase of \( \frac{\pi}{3} \text{ radians} \). The formula for phase difference related to distance \( \Delta x \) is:
\( \Delta \phi = \frac{2\pi}{\lambda} \Delta x \).
Substituting the values:
\( \Delta x = \frac{(\frac{\pi}{3}) \cdot 0.7 \text{ m}}{2\pi} = 0.1167 \text{ m} \).
This shows that a \( \frac{\pi}{3} \text{ rad} \) phase shift corresponds to a separation of about 0.1167 meters.
Key points to remember:
  • Phase difference \( \Delta \phi \) shows the shift between two points in a wave cycle.
  • Measured in radians (rad).
  • Calculated using the formula \( \Delta \phi = \frac{2\pi}{\lambda} \Delta x \).
Angular Frequency
Angular frequency, denoted as \( \omega \), measures how quickly the wave oscillates in terms of radians per second. The formula for angular frequency is:
\( \omega = 2\pi f \).
For our wave with frequency \(500 \text{ Hz}\), the angular frequency is:
\( \omega = 2\pi (500 \text{ Hz}) = 1000\pi \text{ rad/s} \).
Angular frequency is crucial when dealing with time-related wave properties.
Key points to remember:
  • Angular frequency \( \omega \) indicates the rate of oscillation in radians per second.
  • Measured in radians per second (rad/s).
  • Calculated using \( \omega = 2\pi f \).

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Most popular questions from this chapter

Resonant Frequencies A string that is stretched between fixed supports separated by \(75.0 \mathrm{~cm}\) has resonant frequencies of 420 and \(315 \mathrm{~Hz}\), with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Write the Equation Write the equation for a sinusoidal wave traveling in the negative direction along an \(x\) axis and having an amplitude of \(0.010 \mathrm{~m}\), a frequency of \(550 \mathrm{~Hz}\), and a speed of \(330 \mathrm{~m} / \mathrm{s}\).

What Phase Difference What phase difference between two otherwise identical traveling waves, moving in the same direction along a stretched string, will result in the combined wave having an amplitude \(1.50\) times that of the common amplitude of the two combining waves? Express your answer in (a) degrees, (b) radians, and (c) wavelengths.

Standing Wave A standing wave results from the sum of two transverse waves traveling in opposite directions given by $$ y_{1}=(0.05 \mathrm{~m}) \cos ((\pi \mathrm{rad} / \mathrm{m}) x-(4 \pi \mathrm{rad} / \mathrm{s}) t) $$ and $$ y_{2}=(0.05 \mathrm{~m}) \cos ((\pi \mathrm{rad} / \mathrm{m}) x+(4 \pi \mathrm{rad} / \mathrm{s}) t) $$ (a) What is the smallest positive value of \(x\) that corresponds to a node? (b) At what times during the interval \(0 \leq t \leq 0.50 \mathrm{~s}\) will the particle at \(x=0.00 \mathrm{~m}\) have zero velocity?

Uniform Rope A uniform rope of mass \(m\) and length \(L\) hangs from a ceiling. (a) Show that the speed of a transverse wave on the rope is a function of \(y\), the distance from the lower end, and is given by \(v^{\text {wave }}=\sqrt{g y}\). (b) Show that the time a transverse wave takes to travel the length of the rope is given by \(t=2 \sqrt{L / g}\)
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