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Chapter 16: Problem 73

At time \(t=0\) and at position \(x=0 \mathrm{~m}\) along a string, a traveling sinusoidal wave with an angular frequency of \(440 \mathrm{rad} / \mathrm{s}\) has displacement \(y=+4.5 \mathrm{~mm}\) and transverse velocity \(u=-0.75 \mathrm{~m} / \mathrm{s}\). If the wave has the general form \(y(x, t)=y_{m} \sin (k x-\omega t+\phi)\), what is phase constant \(\phi ?\)

Short Answer

Expert verified
The phase constant \(\phi\) is approximately 1.21 radians.

Step by step solution

01

Identify Known Variables

We know the angular frequency, \(\omega = 440\, \mathrm{rad/s}\), the displacement at \(t=0\) and \(x=0\) is \(y = +4.5\, \mathrm{mm} = 0.0045\, \mathrm{m}\), and the transverse velocity \(u = -0.75\, \mathrm{m/s}\).
02

Write the Displacement Equation

The displacement \(y(x,t)\) of the wave is given by the equation: \[ y(x, t) = y_m \sin(kx - \omega t + \phi). \] For \(x=0\) and \(t=0\), this becomes:\[ y = y_m \sin(\phi). \] Thus, \(0.0045 = y_m \sin(\phi).\)
03

Calculate Transverse Velocity

The transverse velocity is given by the derivative of \(y(x,t)\) with respect to time:\[ u = \frac{\partial y}{\partial t} = -y_m \omega \cos(kx - \omega t + \phi). \]At \(x=0\) and \(t=0\), this becomes:\[ -0.75 = -y_m \omega \cos(\phi). \] Simplifying gives:\[ 0.75 = y_m \cdot 440 \cdot \cos(\phi). \]
04

Solve for Amplitude \(y_m\)

To find \(y_m\), solve from one of the previous equations:\[ y_m = \frac{0.75}{440 \cos(\phi)}. \] Substitute \(y_m\) in the equation \(0.0045 = y_m \sin(\phi)\):\[ 0.0045 = \left( \frac{0.75}{440 \cos(\phi)} \right) \sin(\phi). \]
05

Solve for \(\phi\)

From the equation:\[ 0.0045 = \left( \frac{0.75}{440 \cos(\phi)} \right) \sin(\phi), \]you have:\[ 0.0045 \cdot 440 \cos(\phi) = 0.75 \sin(\phi). \]This simplifies to:\[ \tan(\phi) = \frac{0.0045 \cdot 440}{0.75}. \]Calculate \(\phi\):\[ \phi = \tan^{-1} \left( \frac{0.0045 \cdot 440}{0.75} \right). \]
06

Compute the Result

Calculate the numerical values:\[ \frac{0.0045 \times 440}{0.75} \approx 2.64 \]and thus\[ \phi = \tan^{-1}(2.64) \approx 1.21 \, \text{radians}. \]

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Key Concepts

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

Understanding Angular Frequency
Let's dive into the world of waves with a specific focus on angular frequency, often denoted as \( \omega \). Angular frequency is pivotal when discussing sinusoidal waves, as it quantifies how fast something oscillates or rotates.
  • It represents the rate of change of the phase of a sinusoidal waveform, or in simpler terms, how quickly the wave cycles through its oscillations.
  • Measured in radians per second (rad/s), it provides a direct link between the physical displacement of the wave and the time it takes to complete cycles.
In our exercise, the given angular frequency is \( 440 \, \text{rad/s} \). Essentially, this tells us that the wave completes one full oscillation when the phase advances by \( 2\pi \) radians, and the wave does this at a rate of \( 440 \) radians per second. Understanding this helps us quantify phenomena related to cycles and oscillations, which often go unnoticed behind the mathematical descriptions.
The Concept of Phase Constant
The phase constant, often represented by \( \phi \), is an essential concept when dealing with sinusoidal waves. It explains the initial angle or phase of the wave at time \( t=0 \) and position \( x=0 \).
  • It modifies where the wave starts in its cycle, serving as a convenient shift that fits the wave to initial conditions.
  • In mathematical terms, it is part of the sine function in the displacement equation \( y(x, t) = y_m \sin(kx - \omega t + \phi) \).
Our exercise aims to determine this phase constant by utilizing given values of displacement and transverse velocity. As seen, using the displacement \( y = +4.5 \, \text{mm} \) and the velocity \( u = -0.75\, \text{m/s} \) at \( t=0\). By solving the wave equations, you calculate \( \phi \approx 1.21 \, \text{radians} \). This phase constant ensures that our wave equation accurately reflects the actual starting conditions of the wave on the string.
Exploring Transverse Velocity
Transverse velocity is the speed at which points on the wave move perpendicular to the direction of wave propagation.
The transverse velocity at any point on a wave provides insights into the wave鈥檚 energy and how quickly it displaces.
  • It is derived as the time derivative of displacement, represented as \( \frac{\partial y}{\partial t} = -y_m \omega \cos(kx - \omega t + \phi) \).
  • This velocity is crucial to understand how different points on a wave move as the wave travels along a medium like a string.
In the given problem, the transverse velocity is \( u = -0.75 \, \text{m/s} \) at \( t=0 \) and \( x=0 \). Plugging this into the derived equation allows us to isolate and solve for the phase constant \( \phi \). This relationship showcases the connection between angular movement (\( \omega \)) and spatial displacement within wave mechanics, demonstrating the beautifully intricate behavior of traveling waves.

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Most popular questions from this chapter

Four waves are to be sent along the same string, in the same direction: $$ \begin{array}{l} y_{1}(x, t)=(4.00 \mathrm{~mm}) \sin (2 \pi x-400 \pi t) \\ y_{2}(x, t)=(4.00 \mathrm{~mm}) \sin (2 \pi x-400 \pi t+0.7 \pi) \\ y_{3}(x, t)=(4.00 \mathrm{~mm}) \sin (2 \pi x-400 \pi t+\pi) \\ y_{4}(x, t)=(4.00 \mathrm{~mm}) \sin (2 \pi x-400 \pi t+1.7 \pi) . \end{array} $$ What is the amplitude of the resultant wave?

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?

A sinusoidal transverse wave traveling in the negative direction of an \(x\) axis has an amplitude of \(1.00 \mathrm{~cm}\), a frequency of \(550 \mathrm{~Hz}\), and a speed of \(330 \mathrm{~m} / \mathrm{s}\). If the wave equation is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t)\), what are (a) \(y_{m},(\mathrm{~b}) \omega,(\mathrm{c}) k\), and (d) the correct choice of sign in front of \(\omega\) ?

Use the wave equation to find the speed of a wave given by $$ y(x, t)=(3.00 \mathrm{~mm}) \sin \left[\left(4.00 \mathrm{~m}^{-1}\right) x-\left(7.00 \mathrm{~s}^{-1}\right) t\right] . $$

A transverse sinusoidal wave is generated at one end of a long, horizontal string by a bar that moves up and down through a distance of \(1.00 \mathrm{~cm} .\) The motion is continuous and is repeated regularly 120 times per second. The string has linear density \(120 \mathrm{~g} / \mathrm{m}\) and is kept under a tension of \(90.0 \mathrm{~N}\). Find the maximum value of (a) the transverse speed \(u\) and (b) the transverse component of the tension \(\tau\). (c) Show that the two maximum values calculated above occur at the same phase values for the wave. What is the transverse displacement \(y\) of the string at these phases? (d) What is the maximum rate of energy transfer along the string? (e) What is the transverse displacement \(y\) when this maximum transfer occurs? (f) What is the minimum rate of energy transfer along the string? (g) What is the transverse displacement \(y\) when this minimum transfer occurs?
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