/*! 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 38 A forced damped spring-mass syst... [FREE SOLUTION] | 91Ó°ÊÓ

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A forced damped spring-mass system (Fig. \(\mathrm{P} 28.38\) ) has the following ordinary differential equation of motion: \\[ m \frac{d^{2} x}{d t^{2}}+a\left|\frac{d x}{d t}\right| \frac{d x}{d t}+k x=F_{o} \sin (\omega t) \\] where \(x=\) displacement from the equilibrium position, \(t=\) time, \(m=2 \mathrm{kg}\) mass \(, a=5 \mathrm{N} /(\mathrm{m} / \mathrm{s})^{2}\), and \(k=6 \mathrm{N} /(\mathrm{m} / \mathrm{s})^{2}\). The damping term is nonlinear and represents air damping. The forcing function \(F_{0} \sin (\omega t)\) has values of \(F_{o}=2.5 \mathrm{N}\) and \(\omega=0.5 \mathrm{rad} / \mathrm{sec} .\) The initial conditions are Initial velocity' \\[ \frac{d x}{d t}=0 \mathrm{m} / \mathrm{s} \\] Initial displacement \\[ x=1 \mathrm{m} \\] Solve this equation using a numerical method over the time period \(0 \leq t \leq 15\) s. Plot the displacement and velocity versus time, and plot the forcing function on the same curve. Also, develop a separate plot of velocity versus displacement.

Short Answer

Expert verified
To solve the given forced damped spring-mass system with the provided parameters and initial conditions, we first convert the second-order ODE into two first-order ODEs: \( \frac{dx}{dt} = y\) \( \frac{dy}{dt} = -\frac{a}{m} |y| y - \frac{k}{m}x + \frac{F_0}{m} \sin(\omega t)\) Then, using the 4th order Runge-Kutta method, we numerically solve the ODEs over the time period \(0 \leq t \leq 15s\) with a time step size of 0.01s. Finally, we create three plots showing the displacement x(t) vs time t, the velocity y(t) vs time t (along with the forcing function \(F_0\sin(\omega t)\)), and the velocity y(t) vs displacement x(t).

Step by step solution

01

Convert the second-order ODE into two first-order ODEs

To do this, we let the velocity (first derivative of x with respect to t) be another function y. Then the system may be described by: \(y = \frac{dx}{dt}\) \(m \frac{d^{2} x}{d t^{2}} = m \frac{dy}{dt} \) Now, we rewrite the original ODE in terms of x, y, and their derivatives: \(m \frac{dy}{dt} + a|y| y + kx = F_0 \sin(\omega t)\)
02

Define the numerical method for solving the ODEs

Now we have two first-order ODEs which can be written as: \( \frac{dx}{dt} = y\) \( \frac{dy}{dt} = -\frac{a}{m} |y| y - \frac{k}{m}x + \frac{F_0}{m} \sin(\omega t)\) The commonly used numerical method is Euler's method. However, to obtain better results we will use the Runge-Kutta method of order 4. Create a function for solving the system of ODEs using the 4th order Runge-Kutta method.
03

Define the given parameters and initial conditions

Given parameters: \(m = 2kg, a = 5N/(m/s)^2, k = 6N/(m/s)^2, F_0 = 2.5N, \omega = 0.5rad/s\) Initial conditions: \(x(0) = 1m, y(0) = v(0) = 0m/s\) Define a time period for the numerical solution: \(0 \leq t \leq 15s\)
04

Solve the ODEs using the defined method

Using Runge-Kutta 4th order method with given parameters and initial conditions, solve for x(t) and y(t) (velocity). Choose a reasonable time step size for the numerical method, e.g., 0.01s.
05

Plot the results according to the requests

1. Create a plot for the displacement x(t) vs time t. 2. Create a plot for the velocity y(t) vs time t. On the same plot, also show the forcing function \(F_0sin(\omega t)\). 3. Create a third plot for the velocity y(t) vs displacement x(t). With these steps, the given forced damped spring-mass system would be solved numerically, and the results can be visualized and analyzed through the generated plots.

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Key Concepts

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

Runge-Kutta Method
The Runge-Kutta method is one of the most popular numerical techniques for solving ordinary differential equations (ODEs). It's particularly useful when dealing with complex systems like the forced damped spring-mass system. Unlike basic methods such as Euler's, Runge-Kutta provides higher accuracy by taking multiple intermediate calculations within a single step.

The 4th Order Runge-Kutta method, often just referred to as RK4, takes four approximations within a time step to estimate the next value. These intermediate steps allow it to handle nonlinear terms effectively, which is crucial for systems with nonlinear damping like ours. The RK4 formulas are:
  • Calculate slope estimates: \(k_1 = f(x_n, y_n)\), \(k_2 = f(x_n + h/2, y_n + k_1/2)\), \(k_3 = f(x_n + h/2, y_n + k_2/2)\), \(k_4 = f(x_n + h, y_n + k_3)\)
  • Update next value: \(y_{n+1} = y_n + (h/6)(k_1 + 2k_2 + 2k_3 + k_4)\)
  • Here, \(h\) is the time step size.
This method is most accurate and stable when small step sizes are used, such as in our exercise where we utilized a step size of 0.01s.
Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations involving derivatives of a function with respect to one independent variable, usually time. These equations are fundamental in depicting how physical processes evolve over time.

In our forced damped spring-mass system, the ODE describes the motion of a mass on a spring as influenced by damping and an external force. The original ODE provided:
  • Second-order ODE: \(m \frac{d^{2} x}{d t^{2}} + a\left|\frac{d x}{d t}\right| \frac{d x}{d t} + k x = F_{o} \sin (\omega t)\)
To solve this, it’s converted to two first-order ODEs, making it easier to apply numerical methods. This conversion involves defining velocity as a separate function, adding dimensionality to the problem. For numerical computation, managing ODEs in first-order form provides more flexibility and simplification.
Forced Damped Spring-Mass System
The forced damped spring-mass system is a classic mechanical model used to understand the dynamics of mechanical oscillators. This involves a mass attached to a spring that experiences both damping forces and an external periodic force.

Key elements of the system:
  • The **mass (m)**: Determines the system's inertia, affecting how it responds to forces.
  • The **spring constant (k)**: Represents the spring's stiffness. A higher value means a stiffer spring that resists displacement.
  • The **damping (a)**: In this case, nonlinear air damping, which dissipates energy and reduces amplitude over time.
  • The **forcing function**: In our model, \(F_0 \sin (\omega t)\), applies an oscillating force to the system.
This system exhibits complex behavior due to the interaction between damping, stiffness, and the external force, making analytical solutions difficult but fascinating for numerical simulations.
Numerical Simulation
Numerical simulation is a powerful approach to solving differential equations that cannot be handled analytically. This involves using computing algorithms to approximate solutions over small time steps, allowing researchers and students to visualize and analyze systems without an exact solution.

For our spring-mass system, a numerical simulation provides insights into system behavior over time. By implementing the 4th order Runge-Kutta method, we resolve the motion equations into displacement and velocity profiles. The results include:
  • **Displacement vs. Time**: Shows how far the mass moves from its equilibrium position.
  • **Velocity vs. Time**: Indicates the rate of position change, helping identify peak speeds.
  • **Velocity vs. Displacement (phase plot)**: Offers insights into system stability and energy dissipation.
These plots enable engineers and scientists to predict system behavior under various conditions, improving design and analytical capabilities in mechanical systems.

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Most popular questions from this chapter

The basic differential equation of the elastic curve for a cantilever beam (Fig. \(\mathrm{P} 28.22\) ) is given as \(E I \frac{d^{2} y}{d x^{2}}=-P(L-x)\) where \(E=\) the modulus of elasticity and \(l=\) the moment of inertia. Solve for the deflection of the beam using a numerical method. The following parameter values apply: \(E=30,000 \mathrm{ksi}, I=800 \mathrm{in}^{4}\) \(P=1\) kip, \(L=10 \mathrm{ft} .\) Compare your numerical results to the analytical solution, \(y=-\frac{P L x^{2}}{2 E I}+\frac{P x^{3}}{6 E I}\)

Although the model in Prob. 28.18 works adequately when population growth is unlimited, it breaks down when factors such as food shortages, pollution, and lack of space inhibit growth. In such cases, the growth rate itself can be thought of as being inversely proportional to population. One model of this relationship is \(G=G^{\prime}\left(p_{\text {nlax }}-p\right)\) where \(G^{\prime}=\) a population-dependent growth rate (per people-year) and \(p_{\max }=\) the maximum sustainable population. Thus, when population is small \(\left(p \ll p_{\max }\right),\) the growth rate will be at a high constant rate of \(G^{\prime} p_{\max } .\) For such cases, growth is unlimited and Eq. (P28.19) is essentially identical to Eq. (P28.18). However, as population grows (that is, \(p\) approaches \(p_{\max }\) ), \(G\) decreases until at \(p=p_{\max }\) it is zero. Thus, the model predicts that, when the population reaches the maximum sustainable level, growth is nonexistent, and the system is at a steady state. Substituting Eq. (P28.19) into Eq. (P28.18) yields \(\frac{d p}{d t}=G^{\prime}\left(p_{\max }-p\right) p\) For the same island studied in Prob. 28.18 , employ Heun's method (without iteration) to predict the population at \(t=20\) years, using a step size of 0.5 year. Employ values of \(G=10^{-5}\) per people-year and \(p_{\max }=20,000\) people. At time \(t=0,\) the island has a population of 6000 people. Plot \(p\) yersus \(t\) and interpret the shape of the curve.

The following ordinary differential equation desdribes the motion of a damped spring-mass system (Fig. P28.37): \(m \frac{d^{2} x}{d t^{2}}+a\left|\frac{d x}{d t}\right| \frac{d x}{d t}+b x^{3}=0\) where \(x=\) displacement from the equilibrium position, \(t=\) time, \(m=1 \mathrm{kg}\) mass, and \(a=5 \mathrm{N} /(\mathrm{m} / \mathrm{s})^{2}\). The damping term is nonlinear and represents air damping. The spring is a cubic spring and is also nonlinear with \(b=5 \mathrm{N} / \mathrm{m}^{3}\) The initial conditions are Initial velocity \(\frac{d x}{d t}=0.5 \mathrm{m} / \mathrm{s}\) Initial displacement \(\quad x=1 \mathrm{m}\) Solve this equation using a numerical method over the time period \(0 \leq t \leq 8\) s. Plot the displacement and velocity versus time and plot the phase-plane portrait (velocity versus displacement) for all the following cases: (a) A similar linear equation \\[ m \frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+5 x=0 \\] (b) The nonlinear equation with only a nonlinear spring term \\[ \frac{d^{2} x}{d t^{2}}+2 \frac{d x}{d t}+b x^{3}=0 \\] (c) The nonlinear equation with only a nonlinear damping term 9 \(m \frac{d^{2} x}{d t^{2}}+a\left|\frac{d x}{d t}\right| \frac{d x}{d t}+5 x=0\) (d) The full nonlinear equation where both the damping and spring at terms are nonlinear e \(\mathrm{a}_{\mathrm{a}} \quad m \frac{d^{2} x}{d t^{2}}+a\left|\frac{d x}{d t}\right| \frac{d x}{d t}+b x^{3}=0\)

A spherical ice cube (an "ice sphere") that is 6 cm in diameter is removed from a \(0^{\circ} \mathrm{C}\) freezer and placed on a mesh screen at room temperature \(T_{a}=20^{\circ} \mathrm{C} .\) What will be the diameter of the ice cube as a function of time out of the freezer (assuming that all the water that has melted immediately drips through the screen)? The heat transfer coefficient \(h\) for a sphere in a still room is about \(3 \mathrm{W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ). The heat flux from the ice sphere to the air is given by $$\mathrm{Flux}=\frac{q}{A}=h\left(T_{a}-T\right)$$ where \(q=\) heat and \(A=\) surface area of the sphere. Use a numerical method to make your calculation. Note that the latent heat of fusion is \(333 \mathrm{kJ} / \mathrm{kg}\) and the density of ice is approximately \(0.917 \mathrm{kg} / \mathrm{m}^{3}\)

The following equation can be used to model the deflection of a sailboat mast subject to a wind force: \(\frac{d^{2} y}{d z^{2}}=\frac{f}{2 E I}(L-z)^{2}\) where \(f=\) wind force, \(E=\) modulus of elasticity, \(L=\) mast and \(I=\) moment of inertia. Calculate the deflection if \(y=0\) and \(d y / d z=0\) at \(z=0 .\) Use parameter values of \(f=60, L=30, E=\) \(1.25 \times 10^{8},\) and \(I=0.05\) for your computation.

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