91Ó°ÊÓ

CALC Equation \((15.7)\) for a sinusoidal wave can be made more general by including a phase angle \(\phi,\) where \(0 \leq \phi \leq 2 \pi(\) in radians). Then the wave function \(y(x, t)\) becomes $$y(x, t)=A \cos (k x-\omega t+\phi)$$ (a) Sketch the wave as a function of \(x\) at \(t=0\) for \(\phi=0,\) \(\phi=\pi / 4, \phi=\pi / 2, \phi=3 \pi / 4,\) and \(\phi=3 \pi / 2 .\) (b) Calculate the transverse velocity \(v_{y}=\partial y / \partial t .\) (c) At \(t=0,\) a particle on the string at \(x=0\) has displacement \(y=A / \sqrt{2} .\) Is this enough information to determine the value of \(\boldsymbol{\phi} ?\) In addition, if you are told that a particle at \(x=0\) is moving toward \(y=0\) at \(t=0,\) what is the value of \(\phi (\mathrm{d})\) Explain in general what you must know about the wave's behavior at a given instant to determine the value of \(\phi\) .

Step by step solution

01

Sketching the Wave Function

The general wave function is given as \( y(x, t) = A \cos(kx - \omega t + \phi) \). At \( t = 0 \), the function becomes \( y(x, 0) = A \cos(kx + \phi) \). For various phase angles, the sketches are as follows:- For \( \phi = 0 \), the wave starts at the maximum amplitude, \( A \).- For \( \phi = \frac{\pi}{4} \), the wave is shifted a quarter-period to the left relative to \( \phi = 0 \).- For \( \phi = \frac{\pi}{2} \), the wave crosses zero amplitude ((cosine zero)) shifted left by a quarter-period.- For \( \phi = \frac{3\pi}{4} \), wave peaks are further shifted left compared to \( \phi = 0 \).- For \( \phi = \frac{3\pi}{2} \), the wave is again back to crossing zero, but opposite in phase.Visualize and sketch each scenario to understand the phase shifts.

02

Calculating the Transverse Velocity

The transverse velocity is calculated via the derivative: \[ v_y = \frac{\partial y}{\partial t} = A \omega \sin(kx - \omega t + \phi) \].This measures how fast the string moves perpendicular to the wave's propagation direction.

03

Analyzing Displacement for Particle at x=0

Given: At \( t=0 \), \( y = \frac{A}{\sqrt{2}} \) at \( x = 0 \). The equation becomes \[ \frac{A}{\sqrt{2}} = A \cos(\phi) \], which simplifies to \[ \cos(\phi) = \frac{1}{\sqrt{2}} \].Solving gives \( \phi = \frac{\pi}{4} \) or \( \phi = \frac{7\pi}{4} \).

04

Determining Particle Motion Direction

Given that at \( x = 0 \), the particle is moving towards \( y = 0 \) at \( t = 0 \), it implies that \( v_y < 0 \).Using previously calculated \( v_y = A \omega \sin(\phi) \), for \( v_y < 0 \), \( \sin(\phi) < 0 \). Thus \( \phi = \frac{7\pi}{4} \) instead of \( \frac{\pi}{4} \).

05

General Conditions to Determine Phase Angle (\phi)

To determine the phase \( \phi \) accurately, both the displacement \( y \) and the direction of motion (velocity sign) must be known at a point (e.g., at \( x=0, t=0 \)). This allows calculating both \( \cos(\phi) \) and \( \sin(\phi) \) to pinpoint the correct angle within the given range.

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Key Concepts

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

Phase Angle

The phase angle \( \phi \) is a crucial aspect of sinusoidal wave equations. It represents the initial angle of the wave at the starting point of measurement. You can think of it as the wave's starting point on its journey through time and space. Introducing a phase angle can help us adjust the wave to match specific initial conditions.
At \( t = 0 \), the wave equation simplifies to \( y(x, 0) = A \cos(kx + \phi) \). Here, the phase angle shifts the wave along the x-axis without changing its shape or amplitude. Each value of \( \phi \) rotates, or shifts, the wave to the left or right:

• For \( \phi = 0 \), the wave starts at maximum amplitude.
• When \( \phi = \pi/4 \), the wave is shifted a quarter of its period to the left compared to \( \phi = 0 \).
• \( \phi = \pi/2 \) moves the wave crossing zero amplitude at a different position.
• As \( \phi \) increases, so does the horizontal shift, with \( \phi = 3\pi/2 \) positioning the wave oppositely, at zero amplitude but in opposite phase to \( \phi = 0 \).

To find the exact phase angle, we might use both amplitude and movement direction of a point on the wave, allowing us to calculate both \( \cos(\phi) \) and \( \sin(\phi) \). This will ensure the waveform reflects the real wave behaviors precisely.

Wave Function

A wave function helps describe the characteristics of a wave as it travels through space and time. In this case, the sinusoidal wave function is given by \( y(x, t) = A \cos(kx - \omega t + \phi) \). This function tells us all about the wave's profile by revealing its displacement at any point \( x \) and time \( t \).
The components of this wave function include:

• Amplitude \( A \), which represents the maximum displacement from rest.
• Wave number \( k \), defining how many wavelengths fit into a unit distance, calculated as \( k = \frac{2\pi}{\lambda} \).
• Angular frequency \( \omega \), indicating how many cycles occur in a unit of time, calculated as \( \omega = 2\pi f \), where \( f \) is the frequency.

By adjusting the phase angle \( \phi \), the function can accurately reflect different starting points and help visualize how waves move.
When graphed, this function produces a smooth curve that continuously oscillates between positive and negative amplitudes, mirroring the familiar shape of waves we see in water or sound. Observing the wave function at different times provides insights into the wave's propagation and how different parameters affect its shape and behavior.

Transverse Velocity

Transverse velocity is a vital concept when analyzing wave motion, as it measures how fast a point on the wave moves perpendicular to the direction of wave propagation. For our sinusoidal wave function, it is defined as the partial derivative of the wave function with respect to time, \( v_y = \frac{\partial y}{\partial t} \).
This calculation results in the transverse velocity formula: \( v_y = A \omega \sin(kx - \omega t + \phi) \).

Here's what you need to know about transverse velocity:

• The amplitude \( A \) and angular frequency \( \omega \) both play a pivotal role in dictating the maximum possible speed of wave points.
• The sine function indicates that this velocity varies in a sinusoidal pattern itself, peaking at \( A \omega \) when the sine term equals \( \,\pm 1 \).
• This velocity is essential for determining how energy moves through the wave, affecting phenomena like sound transmission or ripples on water.

Understanding the behavior of transverse velocity can also guide us in predicting how fast disturbances or energy will move along a string or medium, helping in real-world applications like musical instruments and engineering tasks.