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Chapter 16: Problem 127

A sinusoidal wave travels down a taut, horizontal string with a linear mass density of \(\mu=0.060 \mathrm{kg} / \mathrm{m} .\) The magnitude of maximum vertical acceleration of the wave is \(a_{y \max }=0.90 \mathrm{cm} / \mathrm{s}^{2}\) and the amplitude of the wave is \(0.40 \mathrm{m} .\) The string is under a tension of \(F_{T}=600.00 \mathrm{N}\) The wave moves in the negative \(x\) -direction. Write an equation to model the wave.

Short Answer

Expert verified
The equation for the sinusoidal wave traveling down the string is: \[y(x, t) = 0.40 \sin(0.0015x - 0.15t + \phi)\]

Step by step solution

01

Find the wave speed (v)

Given the tension (FT) and linear mass density (渭), we can find the wave speed (v) using the formula: \[v = \sqrt{\frac{F_{T}}{\mu}}\] Plug in the given values: \[v = \sqrt{\frac{600.00 \, \mathrm{N}}{0.060 \, \mathrm{kg/m}}}\] Calculate the wave speed: \[v = \sqrt{10000 \, \mathrm{m^2/s^2}} = 100 \, \mathrm{m/s}\]
02

Find the angular frequency (蠅)

We are given the maximum vertical acceleration (ay_max) and amplitude (A). The formula linking maximum vertical acceleration, amplitude, and angular frequency is: \[a_{y \max} = A \cdot \omega^2\] Solve for 蠅: \[\omega = \sqrt{\frac{a_{y \max}}{A}}\] Plug in the given values: \[\omega = \sqrt{\frac{0.90 \, \mathrm{cm/s^2}}{0.40 \, \mathrm{m}}}\] Convert cm/s虏 to m/s虏: \[0.90 \, \mathrm{cm/s^2} \cdot \frac{1 \, \mathrm{m}}{100 \, \mathrm{cm}} = 0.009 \, \mathrm{m/s^2}\] Calculate the angular frequency: \[\omega = \sqrt{\frac{0.009 \, \mathrm{m/s^2}}{0.40 \, \mathrm{m}}} = 0.15 \, \mathrm{rad/s}\]
03

Find the wave number (k)

We have the wave speed (v) and the angular frequency (蠅). We can find the wave number (k) using the formula: \[v = \frac{\omega}{k}\] Solve for k: \[k = \frac{\omega}{v}\] Plug in the calculated values: \[k = \frac{0.15 \, \mathrm{rad/s}}{100 \, \mathrm{m/s}}\] Calculate the wave number: \[k = 0.0015 \, \mathrm{rad/m}\]
04

Write the equation of the wave

Now that we have the amplitude (A), wave number (k), angular frequency (蠅), and knowing the wave moves in the negative x-direction, we can write the equation of the wave using the general equation for sinusoidal waves: \[y(x, t) = A \cdot \sin(kx \pm \omega t + \phi)\] Since the wave is traveling in the negative x-direction, we will use the minus sign: \[y(x, t) = A \cdot \sin(kx - \omega t + \phi)\] Plug in the given and calculated values: \[y(x, t) = 0.40 \, \mathrm{m} \cdot \sin(0.0015 \, \mathrm{rad/m} \cdot x - 0.15 \, \mathrm{rad/s} \cdot t + \phi)\] As we don't have any information to determine the phase constant (蠁), we will leave it as a variable in the equation: \[y(x, t) = 0.40 \, \mathrm{m} \cdot \sin(0.0015x - 0.15t + \phi)\] The final equation for the sinusoidal wave is: \[y(x, t) = 0.40 \sin(0.0015x - 0.15t + \phi)\]

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Key Concepts

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

Wave Speed Calculation
To understand sinusoidal waves traveling along a string, it's essential to calculate the wave speed, often denoted by the symbol \(v\). The speed of a wave on a string depends on two key elements: the tension in the string \(F_T\) and the string's linear mass density \(\mu\). The formula for wave speed is derived from balancing the forces involved and is given by:
  • \(v = \sqrt{\frac{F_T}{\mu}}\)
In this scenario, we see that the tension \(F_T = 600.00 \, \mathrm{N}\) and the linear mass density \(\mu = 0.060 \, \mathrm{kg/m}\). Substituting these into our formula:
  • \(v = \sqrt{\frac{600.00}{0.060}}\)
This yields \(v = 100 \, \mathrm{m/s}\). Thus, the wave propagates through the string at this speed. Knowledge of wave speed is crucial in understanding how quickly a disturbance travels through a medium.
Angular Frequency
Angular frequency, represented by the Greek letter \(\omega\), is a measure of how fast the wave oscillates as it moves. It's linked to the cycle of the wave and is measured in radians per second. In sinusoidal waves, knowing the angular frequency allows us to understand how fast the particles in the wave are oscillating up and down.
  • The formula to find \(\omega\) is linked to the maximum vertical acceleration \(a_{y \max}\) and amplitude \(A\).
  • \(a_{y \max} = A \cdot \omega^2\)
Given that the maximum vertical acceleration \(a_{y \max} = 0.90 \, \mathrm{cm/s^2}\) and the amplitude \(A = 0.40 \, \mathrm{m}\), we first convert the acceleration to \(\mathrm{m/s^2}\) resulting in \(0.009 \, \mathrm{m/s^2}\). Solving for \(\omega\):
  • \(\omega = \sqrt{\frac{0.009}{0.40}}\)
After calculation, we find \(\omega = 0.15 \, \mathrm{rad/s}\). This value indicates the speed of oscillation in terms of angle, a crucial component for forming our wave equation.
Wave Equation Modeling
Modeling a wave accurately using an equation is the final step, combining all calculated values and known properties. The structure used for describing sinusoidal waves mathematically is generally:
  • \(y(x, t) = A \cdot \sin(kx \pm \omega t + \phi)\)
For this, we need the amplitude \(A\), wave number \(k\), angular frequency \(\omega\), and phase constant \(\phi\). Given that the wave moves in the negative \(x\)-direction, we choose the minus sign in the equation. Here is how each parameter is determined:
  • Amplitude \(A\) is given as \(0.40 \, \mathrm{m}\).
  • The wave number \(k\) is found using \(k = \frac{\omega}{v}\), resulting in \( k = 0.0015 \, \mathrm{rad/m}\).
  • Angular frequency \(\omega\) is \(0.15 \, \mathrm{rad/s}\).
Thus, the wave equation is structured as:
  • \(y(x, t) = 0.40 \cdot \sin(0.0015x - 0.15t + \phi)\)
The phase constant \(\phi\) remains undetermined in this scenario, so it is left as a variable. This comprehensive equation gives a complete representation of how the sinusoidal wave behaves as it travels along the string.

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Most popular questions from this chapter

A sinusoidal, transverse wave is produced on a stretched spring, having a period \(T\). Each section of the spring moves perpendicular to the direction of propagation of the wave, in simple harmonic motion with an amplitude A. Does each section oscillate with the same period as the wave or a different period? If the amplitude of the transverse wave were doubled but the period stays the same, would your answer be the same?

A 2.40-m wire has a mass of 7.50 g and is under a tension of \(160 \mathrm{N}\). The wire is held rigidly at both ends and set into oscillation. (a) What is the speed of waves on the wire? The string is driven into resonance by a frequency that produces a standing wave with a wavelength equal to 1.20 m. (b) What is the frequency used to drive the string into resonance?

What is the wavelength of an earthquake that shakes you with a frequency of \(10.0 \mathrm{Hz}\) and gets to another city 84.0 km away in 12.0 s?

A copper wire has a density of \(\rho=8920 \mathrm{kg} / \mathrm{m}^{3},\) a radius of \(1.20 \mathrm{mm},\) and a length \(L .\) The wire is held under a tension of \(10.00 \mathrm{N}\). Transverse waves are sent down the wire. (a) What is the linear mass density of the wire? (b) What is the speed of the waves through the wire?

A tuning fork vibrates producing sound at a frequency of \(512 \mathrm{Hz}\). The speed of sound of sound in air is \(v=343.00 \mathrm{m} / \mathrm{s}\) if the air is at a temperature of \(20.00^{\circ} \mathrm{C}\).What is the wavelength of the sound?
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