Problem 68
Question
Consider Problem \(13.52\) again, in which the central-difference method (Eq. 13.10-5) is applied to a spring-mass system for which \(k=m=\omega=1\). Now use \(\Delta t=\sqrt{3.96}\) and start the algorithm using \(u_{0}=0\) and \(u_{-1}=-1\). Follow the motion for at least ten cycles, and observe that the computed amplitude displays "beating" but no net growth.
Step-by-Step Solution
Verified Answer
The amplitudes computed using the central-difference method over ten cycles in a spring-mass system display 'beating' phenomenon - a fluctuation in amplitudes, but no significant growth or decay in the amplitude over time. This is consistent with the properties of harmonic motion. The beating pattern can be seen more clearly by preserving the computed amplitudes and plotting them over the ten cycles.
1Step 1: Understanding the central-difference method
The central-difference method deals with the approximate solution of ordinary differential equations. The equation for this method is generally given as \(u_{i+1}=2u_{i}-u_{i-1}+\Delta t^2 f(t,u)\), where \(f(t,u)\) is the derivative of \(u\) with respect to \(t\). In the case of a spring-mass system, this becomes \(u_{i+1}=2u_{i}-u_{i-1}+\Delta t^2 (-\omega^2 u_{i})\).
2Step 2: Apply the given values
Now we plug our initial conditions \(u_{0}=0, u_{-1}=-1\) and our given values \(k=m=\omega=1\), \(\Delta t=\sqrt{3.96}\) into the formula. This gives us \(u_{1}=2(0)-(-1)+(\sqrt{3.96})^2*(-1*0) = 1\).
3Step 3: Perform iteration for ten cycles
Repeat the calculation by updating the indices i.e., \(i=i+1\). We will calculate \(u_{2}\), \(u_{3}\), ... up to \(u_{10}\) at least, to follow the motion for ten cycles.
4Step 4: Observe the 'beating' and net growth
'Beating' refers to the variation in amplitude over the ten cycles. Although the amplitude might fluctuate, there should be no net growth which means initial and final amplitudes (after ten cycles) should ideally be the same or very close.
Key Concepts
Ordinary Differential EquationsSpring-Mass SystemNumerical Analysis
Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations involving a function and its derivatives. They are pivotal in describing various phenomena in physics, engineering, and other sciences. An ODE contains one or more functions and their derivatives with respect to one variable. In many situations, they model how a system evolves over time.
ODEs are classified based on their order, which is determined by the highest derivative present in the equation. For example, a first-order ODE involves only the first derivative of the function, while a second-order ODE involves the second derivative. These equations may be linear or non-linear, homogeneous or non-homogeneous, depending on their form.
A common example of an ODE is the equation governing the motion of a spring. This involves a second-order differential equation because it relates the second derivative of displacement (acceleration) to the forces acting on the spring. Solving ODEs analytically provides a formula for calculating system behavior at any time point, although many complex ODEs must be solved using numerical methods.
ODEs are classified based on their order, which is determined by the highest derivative present in the equation. For example, a first-order ODE involves only the first derivative of the function, while a second-order ODE involves the second derivative. These equations may be linear or non-linear, homogeneous or non-homogeneous, depending on their form.
A common example of an ODE is the equation governing the motion of a spring. This involves a second-order differential equation because it relates the second derivative of displacement (acceleration) to the forces acting on the spring. Solving ODEs analytically provides a formula for calculating system behavior at any time point, although many complex ODEs must be solved using numerical methods.
Spring-Mass System
A spring-mass system is a fundamental physical system often studied in mechanics. It comprises a mass attached to a spring, which may be stretched or compressed. The system is characterized by parameters such as stiffness constant (\(k\)) and mass (\(m\)). These parameters influence how the system reacts to external forces.
Spring-mass systems are governed by Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. This can be expressed mathematically by the equation: \[ F = -kx \], where \( F \) is the force, \( k \) is the stiffness constant, and \( x \) is the displacement from equilibrium.
When combined with Newton's Second Law, \( F = ma \), we derive the second-order differential equation: \[ m\frac{d^2x}{dt^2} = -kx \]. In the context of a spring-mass system, this differential equation describes how the position of the mass changes over time when it is attached to an ideal spring. Analyzing this equation helps us understand the natural frequency and dynamics of the vibration of the system.
Spring-mass systems are governed by Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from its equilibrium position. This can be expressed mathematically by the equation: \[ F = -kx \], where \( F \) is the force, \( k \) is the stiffness constant, and \( x \) is the displacement from equilibrium.
When combined with Newton's Second Law, \( F = ma \), we derive the second-order differential equation: \[ m\frac{d^2x}{dt^2} = -kx \]. In the context of a spring-mass system, this differential equation describes how the position of the mass changes over time when it is attached to an ideal spring. Analyzing this equation helps us understand the natural frequency and dynamics of the vibration of the system.
Numerical Analysis
Numerical Analysis involves using algorithms to approximate solutions for mathematical problems that cannot be solved analytically. This is especially useful for complex Ordinary Differential Equations (ODEs), where exact solutions might be difficult or impossible to find. Techniques in numerical analysis are deployed to simulate and understand the behavior of physical systems such as the spring-mass system.
The central-difference method is a popular numerical technique employed to solve ODEs. It provides approximations by utilizing discrete values of the function over a mesh of points. For the spring-mass system example, the central-difference method involves a specific formula that includes previous and current values with the time step (\( \Delta t \)). It is given by: \[ u_{i+1}=2u_{i}-u_{i-1}+\Delta t^2(-\omega^2 u_{i}) \].
Through numerical analysis, particularly the central-difference technique, students can explore properties like 'beating' in dynamic systems and how initial conditions impact the system's evolution.
The central-difference method is a popular numerical technique employed to solve ODEs. It provides approximations by utilizing discrete values of the function over a mesh of points. For the spring-mass system example, the central-difference method involves a specific formula that includes previous and current values with the time step (\( \Delta t \)). It is given by: \[ u_{i+1}=2u_{i}-u_{i-1}+\Delta t^2(-\omega^2 u_{i}) \].
- \(u_{i+1}\) represents the next approximation.
- \(u_{i}\) represents the current value.
- \(u_{i-1}\) represents the previous value.
- \(\omega\) is the natural frequency of the system.
Through numerical analysis, particularly the central-difference technique, students can explore properties like 'beating' in dynamic systems and how initial conditions impact the system's evolution.
Other exercises in this chapter
Problem 52
A particle of unit mass is supported by a spring of unit stiffness. There is no damping and no external load. Thus \(k=m=\omega=1\). At time \(t=0\), the partic
View solution Problem 53
A particle of unit mass is supported by a spring of unit stiffness. There is no damping and no external load. Thus \(k=m=\omega=1\). At time \(t=0\), the partic
View solution Problem 73
The uncoupled equations produced by a modal analysis (Section 13.6) have a lower \(\omega_{\max }\) than \(\omega_{\max }\) of the full system. Hence, in integr
View solution Problem 46
Model a simply supported beam by a single element. Let \(L=1.0 \mathrm{~m}, A=\) \(0.0002 \mathrm{~m}^{2}, E I=300.0 \mathrm{~N} \cdot \mathrm{m}^{2}\), and \(\
View solution