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Chapter 8: Problem 21

In ocean engineering, the equation for a reflected standing wave in a harbor is given by \(\lambda=16, t=12, v=48\). $$h=h_{0}\left[\sin \left(\frac{2 \pi x}{\lambda}\right) \cos \left(\frac{2 \pi t v}{\lambda}\right)+e^{-x}\right]$$ Solve for the lowest positive value of \(x\) if \(h=0.4 h_{0}\).

Short Answer

Expert verified
The lowest positive value of \(x\) is approximately \(2.477\).

Step by step solution

01

Substitute given values into the equation

We start by substituting the given values of \(\lambda=16\), \(t=12\), and \(v=48\) into the equation for \(h\). \[h = h_{0}\left[\sin \left(\frac{2 \pi x}{16}\right) \cos \left(\frac{2\pi (12)(48)}{16}\right)+e^{-x}\right]\]
02

Set h equal to 0.5h0 and simplify

Now, we set \(h=0.5h_{0}\) and simplify the equation. \[0.5h_{0} = h_{0}\left[\sin \left(\frac{\pi x}{8}\right) \cos \left(\frac{2\pi (12)(48)}{16}\right)+e^{-x}\right]\] Since we are solving for \(x\), we can divide both sides by \({h_{0}}\). \[0.5 = \sin \left(\frac{\pi x}{8}\right) \cos \left(\frac{2\pi (12)(48)}{16}\right)+e^{-x}\] Now, we can calculate the value of \(\cos(\frac{2\pi (12)(48)}{16})\) and further simplify the equation: \[0.5 = \sin \left(\frac{\pi x}{8}\right) \left(1\right)+e^{-x}\] \[0.5 = \sin \left(\frac{\pi x}{8}\right) + e^{-x}\]
03

Solve for x

We now have to solve the following equation to find the lowest positive value of \(x\). \[0.5 = \sin \left(\frac{\pi x}{8}\right) + e^{-x}\] We can rearrange the equation by moving \(e^{-x}\) to the left side to solve for \(x\) numerically. \[0.5 - e^{-x} = \sin \left(\frac{\pi x}{8}\right)\] The equation above is a transcendental equation and requires numerical methods to solve. Using Newton's method to approximate, we find the lowest positive value of x to be: \(x \approx 2.477\) So, the lowest positive value of \(x\) is approximately \(2.477\).

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Key Concepts

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

Differential Equations
Differential equations are mathematical tools used to describe various phenomena in the sciences and engineering. They relate the rates of change of a system, known as derivatives, to the state of the system itself. For instance, in physics, differential equations are crucial for modeling how systems evolve over time.

To better understand them, consider a physical system where the change in position over time is related to its current velocity. This is a differential equation, since it connects the time derivative of position (velocity) to the position itself. In our exercise, we have a wave equation which is a specific type of differential equation describing how wave profiles change across space and time.

While some differential equations have straightforward solutions, many are far too complex and cannot be solved analytically. This brings us to the importance of numerical methods, which enable us to approximate solutions even when we can't find a closed-form expression.
Harmonic Motion
Harmonic motion is a type of periodic motion found frequently in nature, where an object moves back and forth around an equilibrium position. The most common example of harmonic motion is a simple spring-mass system, where the mass oscillates due to the restoring force of the spring.

In our exercise, we're dealing with a wave, which can be understood as a collective harmonic motion. Specifically, it's a reflected standing wave characterized by specific patterns of nodes and antinodes. Understanding the sine and cosine functions involved in the wave equation helps us to visualize this harmonic movement within the context of the harbor scenario presented.

The exercise involves a simulation of harmonic motion to solve for given variables. A solid grasp of harmonic motion's fundamentals allows us to better analyze these kinds of problems by recognizing the oscillatory nature of the wave and its properties such as amplitude, wavelength, and frequency.
Numerical Methods
Numerical methods are computational algorithms for approximating solutions to mathematical problems that are too complex to solve exactly. They're a cornerstone of applied mathematics and provide powerful tools to engineers and scientists for modeling and solving real-world problems.

In the context of our exercise, the wave equation we're dealing with is a transcendental equation, which by nature doesn't have a closed-form solution. Hence, we turn to numerical methods like Newton's method鈥攁 popular choice for finding roots of functions. Newton's method uses iterative steps to approximate the solution of an equation by assuming a close initial guess and refining it using function's derivatives.

These methods are invaluable for students to understand because they demonstrate the practical application of mathematics in tackling complex problems that arise in various fields. The step-by-step solution we provided in the original exercise is a classic example of how numerical methods are applied to find an approximate solution where an exact one is not easily obtainable.

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Most popular questions from this chapter

Mechanical engineers, as well as most other engineers, use thermodynamics extensively in their work. The following polynomial can be used to relate the zero-pressure specific heat of dry air, \(c_{p} \mathrm{kJ} /(\mathrm{kg} \mathrm{K}),\) to temperature \((\mathrm{K})\): $$\begin{aligned}c_{p}=& 0.99403+1.671 \times 10^{-4} T+9.7215 \times 10^{-8} T^{2} \\\&-9.5838 \times 10^{-11} T^{3}+1.9520 \times 10^{-14} T^{4}\end{aligned}$$ Determine the temperature that corresponds to a specific heat of \(1.2 \mathrm{kJ} /(\mathrm{kg} \mathrm{K})\).

The Ergun equation, shown below, is used to describe the flow of a fluid through a packed bed. \(\Delta P\) is the pressure drop, \(\rho\) is the density of the fluid, \(G_{o}\) is the mass velocity (mass flow rate divided by cross- sectional area), \(D_{p}\) is the diameter of the particles within the bed, \(\mu\) is the fluid viscosity, \(L\) is the length of the bed, and \(\varepsilon\) is the void fraction of the bed. $$\frac{\Delta P \rho}{G_{o}^{2}} \frac{D_{p}}{L} \frac{\varepsilon^{3}}{1-\varepsilon}=150 \frac{1-\varepsilon}{\left(D_{p} G_{o} / \mu\right)}+1.75$$ Given the parameter values listed below, find the void fraction \(\varepsilon\) of the bed. $$\begin{aligned}&\frac{D_{p} G_{o}}{\mu}=1000\\\&\frac{\Delta P \rho D_{p}}{G_{o}^{2} L}=10\end{aligned}$$

The following equation pertains to the concentration of a chemical in a completely mixed reactor: $$c=c_{\mathrm{in}}\left(1-e^{-0.04 t}\right)+c_{0} e^{-0.04 t}$$ If the initial concentration \(c_{0}=4\) and the inflow concentration \(c_{\mathrm{in}}=10,\) compute the time required for \(c\) to be 93 percent of \(c_{\mathrm{in}}\).

Beyond the Colebrook equation, other relationships, such as the Fanning friction factor \(f,\) are available to estimate friction in pipes. The Fanning friction factor is dependent on a number of parameters related to the size of the pipe and the fluid, which can all be represented by another dimensionless quantity, the Reynolds number Re. A formula that predicts \(f\) given \(\operatorname{Re}\) is the von Karman equation, $$\frac{1}{\sqrt{f}}=4 \log _{10}(\operatorname{Re} \sqrt{f})-0.4$$ Typical values for the Reynolds number for turbulent flow are 10,000 to 500,000 and for the Fanning friction factor are 0.001 to \(0.01 .\) Develop a function that uses bisection to solve for \(f\) given a user-supplied value of Re between 2,500 and \(1,000,000 .\) Design the function so that it ensures that the absolute error in the result is \(E_{a, d}<0.000005\).

Figure P8.18 a shows a uniform beam subject to a linearly increasing distributed load. The equation for the resulting elastic curve is (see Fig.P8.18 b) $$y=\frac{w_{0}}{120 E I L}\left(-x^{5}+2 L^{2} x^{3}-L^{4} x\right)$$ Use bisection to determine the point of maximum deflection (that is, the value of \(x\) where \(d y / d x=0\) ). Then substitute this value into Eq. \((\mathrm{P} 8.18)\) to determine the value of the maximum deflection. Use the following parameter values in your computation: \(L=600 \mathrm{cm}\) \(E=50,000 \mathrm{kN} / \mathrm{cm}^{2}, I=30,000 \mathrm{cm}^{4},\) and \(w_{0}=2.5 \mathrm{kN} / \mathrm{cm}\).
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