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Chapter 16: Problem 1

If a wave \(y(x, t)=(6.0 \mathrm{~mm}) \sin (k x+(600 \mathrm{rad} / \mathrm{s}) t+\phi)\) travels along a string, how much time does any given point on the string take to move between displacements \(y=+2.0 \mathrm{~mm}\) and \(y=-2.0 \mathrm{~mm}\) ?

Short Answer

Expert verified
The time is approximately 5.24 milliseconds.

Step by step solution

01

Identify the Given Wave Equation

The wave equation is given as \( y(x, t) = (6.0 \text{ mm}) \sin(k x + 600 \text{ rad/s} \cdot t + \phi) \). Here, the amplitude \( A = 6.0 \text{ mm} \), angular frequency \( \omega = 600 \text{ rad/s} \). We are tasked with finding the time it takes for the displacement to move from \( y = +2.0 \text{ mm} \) to \( y = -2.0 \text{ mm} \).
02

Find the Phase Corresponding to Displacements

For the displacement \( y = 2.0 \text{ mm} \), solve for the phase \( \theta_1 \) in terms of \( t \): \( 2.0 \text{ mm} = 6.0 \text{ mm} \cdot \sin(\theta_1) \). This simplifies to \( \sin(\theta_1) = \frac{1}{3} \). Similarly, for \( y = -2.0 \text{ mm} \), solve for the phase \( \theta_2 \): \( -2.0 \text{ mm} = 6.0 \text{ mm} \cdot \sin(\theta_2) \) leading to \( \sin(\theta_2) = -\frac{1}{3} \).
03

Determine Phase Angles

For \( \sin(\theta_1) = \frac{1}{3} \), we have \( \theta_1 = \arcsin\left(\frac{1}{3}\right) \) and \( \pi - \arcsin\left(\frac{1}{3}\right) \). For \( \sin(\theta_2) = -\frac{1}{3} \), \( \theta_2 = -\arcsin\left(\frac{1}{3}\right) \) and \( \pi + \arcsin\left(\frac{1}{3}\right) \). Choose consecutive angles for the wave moving smoothly from \( +2.0 \text{ mm} \) to \( -2.0 \text{ mm} \).
04

Select Consecutive Phase Angles

Choose consecutive phase angles \( \theta_1 = \arcsin\left(\frac{1}{3}\right) \) and \( \theta_2 = \pi + \arcsin\left(\frac{1}{3}\right) \) that correspond to the required transition from \( +2.0 \text{ mm} \) to \( -2.0 \text{ mm} \).
05

Calculate the Time Interval

Since \( \omega = 600 \text{ rad/s} \), the time difference \( \Delta t \) between these angles is \( \Delta t = \frac{\theta_2 - \theta_1}{\omega} = \frac{\left( \pi + \arcsin\left(\frac{1}{3}\right) \right) - \arcsin\left(\frac{1}{3}\right)}{600} \). Simplifying gives \( \Delta t = \frac{\pi}{600} \). Evaluate \( \Delta t \) using \( \pi \approx 3.1416 \).
06

Solve for Numeric Value of Time Interval

Compute \( \Delta t = \frac{3.1416}{600} \approx 0.00524 \text{ seconds} \) or \( 5.24 \text{ milliseconds} \).

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Key Concepts

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

Amplitude
In the context of wave equations, **Amplitude** is a crucial concept. Simply put, it is the maximum displacement from the wave's equilibrium position. In the given wave equation, the amplitude is expressed as 6.0 millimeters. This means that the maximum height (or depth) that the wave can reach from its central axis is 6.0 millimeters.

  • Amplitude affects the energy of the wave鈥攈igher amplitudes indicate more energy.
  • Mathematically, it is the coefficient in front of the sine function in the wave equation.
Understanding amplitude helps in visualizing how "tall" a wave can extend as it oscillates, resembling peaks and valleys.
Angular Frequency
**Angular Frequency** relates to how quickly the wave oscillates over time. For our wave, the angular frequency is 600 rad/s.

  • Angled frequency is denoted by the Greek letter \( \omega \).
  • It contributes to the calculation of the wave's periodic motion.
Angular frequency is critical as it gives insight into how rapid the wave oscillates. A higher angular frequency signifies that the wave completes more cycles in a given time frame.
Phase Angle
The **Phase Angle** can be a bit tricky but is essential for understanding where exactly the wave is at any given moment in its cycle. This angle shifts the wave forward or backward in time without altering its form.

  • Phase angle changes the starting point of the wave.
  • In our problem, it is represented as \( \phi \) in the wave equation.
Phase angles are particularly important when comparing multiple waves. They tell us how "in-step" (or "out-of-step") these waves might be relative to one another.
Sinusoidal Wave
A **Sinusoidal Wave** is a wave that describes a smooth periodic oscillation. It's one of the fundamental types of wave shapes and is given by the sine function \( \sin(\cdot) \).

  • The wave equation we discuss is a sinusoidal wave equation.
  • Such waves are important because they model various natural phenomena like sound and light.
A sinusoidal wave is pleasingly symmetrical, characterized by its continuous wave of peaks and troughs. Mastering this concept eases the study of more complex wave behaviors.
Time Interval Calculation
**Time Interval Calculation** is pivotal when determining how long it takes for particular events to occur鈥攕uch as reaching specific points on a wave.

In the provided exercise, we calculate the time it takes for a point to go from a displacement of +2.0 mm to -2.0 mm:
  • Understand the change in phase needed for this movement.
  • Apply the formula \( \Delta t = \frac{\Delta \theta}{\omega} \) where \( \Delta \theta \) is the change in phase.
  • Substitute angular frequency and phase angles into the formula.
In the example, we calculated the time interval as approximately 5.24 milliseconds, demonstrating the change in vertical position on the wave within a concise duration.

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Most popular questions from this chapter

In Fig. \(16-42\), a string, tied to a sinusoidal oscillator at \(P\) and running over a support at \(Q\), is stretched by a block of mass \(m\). The separation \(L\) between \(P\) and \(Q\) is \(1.20 \mathrm{~m}\), and the frequency \(f\) of the oscillator is fixed at \(120 \mathrm{~Hz}\). The amplitude of the motion at \(P\) is small enough for that point to be considered a node. A node also exists at \(Q .\) A standing wave appears when the mass of the hanging block is \(286.1 \mathrm{~g}\) or \(447.0 \mathrm{~g} .\) but not for any intermediate mass. What is the linear density of the string?

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 nylon guitar string has a linear density of \(7.20 \mathrm{~g} / \mathrm{m}\) and is under a tension of \(150 \mathrm{~N}\). The fixed supports are distance \(D=90.0 \mathrm{~cm}\) apart. The string is oscillating in the standing wave pattern shown in Fig. 16-39. Calculate the (a) speed, (b) wavelength, and (c) frequency of the traveling waves whose superposition gives this standing wave.

If a transmission line in a cold climate collects ice, the increased diameter tends to cause vortex formation in a passing wind. The air pressure variations in the vortexes tend to cause the line to oscillate (gallop), especially if the frequency of the variations matches a resonant frequency of the line. In long lines, the resonant frequencies are so close that almost any wind speed can set up a resonant mode vigorous enough to pull down support towers or cause the line to short out with an adjacent line. If a transmission line has a length of \(347 \mathrm{~m}\), a linear density of \(3.35 \mathrm{~kg} / \mathrm{m}\), and a tension of \(65.2 \mathrm{MN}\), what are (a) the frequency of the fundamental mode and (b) the frequency difference between successive modes?

A \(120 \mathrm{~cm}\) length of string is stretched between fixed supports. What are the (a) longest, (b) second longest, and (c) third longest wavelength for waves traveling on the string if standing waves are to be set up? (d) Sketch those standing waves.
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