/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 The phase angle \(\phi\) between... [FREE SOLUTION] | 91Ó°ÊÓ

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The phase angle \(\phi\) between the forced vibration caused by the rough road and the motion of the car is given by $$\tan \phi=\frac{2\left(c / c_{c}\right)(\omega / p)}{1-(\omega / p)^{2}}$$ As a mechanical engineer, you would like to know if there are cases where \(\phi=\omega / 3-1 .\) Use the other parameters from the section to set up the equation as a roots problem and solve for \(\omega\).

Short Answer

Expert verified
Substitute the given value of \(\phi = \omega / 3 - 1\) into the equation: \(\tan (\omega / 3 - 1) = \frac{2(c / c_{c})(\omega / p)}{1 - (\omega / p)^2}\) Use the identity \(\tan x = \frac {sin x} {cos x}\) and cross-multiply: \(\sin(\omega / 3 - 1)(1 - (\omega / p)^2) - 2(c / c_{c})(\omega / p) (\cos(\omega / 3 - 1)) = 0\) To solve for \(\omega\), use numerical methods or graph the equation. The x-intercepts give the values of \(\omega\) that satisfy the condition.

Step by step solution

01

Substitute the given value of \(\phi\) into the equation

First, let's substitute the given value of \(\phi = \omega / 3 - 1\) into the equation: \(\tan (\omega / 3 - 1) = \frac{2(c / c_{c})(\omega / p)}{1 - (\omega / p)^2}\)
02

Get rid of the tangent function

To get rid of the tangent function, use its identity: \(\tan x = \frac {sin x} {cos x}\) Now we have: \(\frac{\sin(\omega / 3 - 1)}{\cos(\omega / 3 - 1)} = \frac{2(c / c_{c})(\omega / p)}{1 - (\omega / p)^2}\)
03

Cross-multiply

To remove the fractions, cross-multiply: \(\sin(\omega / 3 - 1)(1 - (\omega / p)^2) = 2(c / c_{c})(\omega / p) \cos(\omega / 3 - 1)\)
04

Rearrange the equation and factor

Rearrange the equation to get everything on one side: \(\sin(\omega / 3 - 1)(1 - (\omega / p)^2) - 2(c / c_{c})(\omega / p) (\cos(\omega / 3 - 1)) = 0\) Now we have a roots problem, and we need to solve for \(\omega\).
05

Solve for \(\omega\)

To solve for \(\omega\), we can either use numerical methods to find the roots of the equation (like the Newton-Raphson method), or we can graph the equation and find the x-intercepts. The x-intercepts will give us the values of \(\omega\) that satisfy our condition. Remember, this exercise may not have an analytical solution since we are dealing with trigonometric functions. Using numerical methods or graphing methods is the way to solve for \(\omega\) in this case.

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Key Concepts

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

Phase Angle in Vibrations
Understanding the phase angle in vibrations is essential in analyzing the behavior of oscillatory systems, such as a car driving on a rough road. Phase angle, denoted as \(\phi\), is the offset between the response of a vibrating system and the excitation that causes the vibration. Think of it as a delay between when the road bumps and when the car's suspension reacts. Mathematically, phase angle connects the input force to the response through the tangent function.In the case of forced vibrations, the phase difference indicates how out of sync the two motions are, which can greatly affect the comfort of the vehicle's occupants and the car's handling. When the system is precisely tuned, the phase angle ensures minimal adverse impact from road irregularities. In practical terms, if an engineer knows the phase angle, they can predict and adjust the system to minimize these effects, ensuring a smoother ride.
Tangent Function in Oscillations
The tangent function plays a critical role in analyzing oscillatory motion. The tangent of an angle in a right triangle represents the ratio of the side opposite to the angle to the side adjacent to it. In the context of oscillations, the tangent function helps us express the out-of-phase relationship between two wave-like motions, such as a car's forced vibration from a rough road and its natural oscillation.By using the identity \(\tan x = \frac{\sin x}{\cos x}\), engineers can relate the sine and cosine functions, which are fundamental to describing oscillatory motion. This trigonometric function allows for converting the phase relationship into a solvable equation when examining the behavior of dynamic systems under forced vibration. With this application, handling complex vibrations becomes more manageable, enabling solutions that optimize performance and comfort.
Differential Equations in Engineering
Differential equations are the cornerstone of engineering analysis, providing the means to describe how physical systems evolve over time. These equations relate a function to its derivatives, which represent rates of change. In mechanical engineering, they model the dynamics of systems like a car moving on an uneven road, where forces and motion are continuously changing.For example, when a car experiences forced vibrations due to road roughness, a differential equation can describe the relationship between displacement, velocity, acceleration, and the excitation from the road. Solving such equations, either analytically or numerically, allows engineers to predict system behavior under various conditions and design accordingly. As a result, mastering differential equations unlocks the ability to control and optimize complex systems within the constraints of the real world.
Numerical Methods for Root Finding
Numerical methods for root finding are essential algorithms that enable engineers to find solutions to equations where analytical methods are impractical or impossible. These methods provide an approximate solution to the places where a function crosses the axis, known as roots. In forced vibration problems, such as determining the frequency of oscillation \(\omega\) for a car suspension system, numerical methods like the Newton-Raphson or bisection methods are commonly used.These iterative methods start with an initial guess and refine the solution through repeated calculations. They are particularly useful when dealing with non-linear equations involving trigonometric functions where direct solutions are difficult to ascertain. Numerical root-finding allows for the precise adjustment of system parameters to achieve the desired operational behavior, making it a powerful tool in an engineer's arsenal.

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Most popular questions from this chapter

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 \(E_{a, d}<0.000005\).

Perform the same computation as in Sec. \(8.3,\) but determine the value of \(C\) required for the circuit to dissipate to \(1 \%\) of its original value in \(t=0.05 \mathrm{s},\) given \(R=280 \Omega,\) and \(L=7.5 \mathrm{H} .\) Use (a) a graphical approach, (b) bisection, and (c) root location software such as the Excel Solver, the MATLAB function f zero, or the Mathcad function root.

The concentration of pollutant bacteria \(c\) in a lake decreases according to $$c=75 e^{-1.5 t}+20 e^{-0.075 t}$$ Determine the time required for the bacteria concentration to be reduced to 15 using (a) the graphical method and (b) using the Newton-Raphson method with an initial guess of \(t=6\) and a stopping criterion of \(0.5 \% .\) Check your result.

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. $$\begin{aligned}&\frac{D_{p} G_{o}}{\mu}=1000\\\&\frac{\Delta P \rho D_{p}}{G_{o}^{2} L}=10\end{aligned}$$

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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