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Determine the overshoot of the system of Problem \(2.26\) if it is critically damped and \(v_{0}=-2 \mathrm{~m} / \mathrm{sec}\).

Short Answer

Expert verified
Answer: The overshoot of the critically damped system with an initial velocity of -2 m/s can be expressed as ((A - 2t)e^{-ω₀t_max} - A), where A is the initial displacement, t_max is the time at maximum displacement, and ω₀ is the angular frequency. Further information about the system parameters is required to compute the exact value of the overshoot.

Step by step solution

01

Understand the properties of a critically damped system

A critically damped system is a second-order differential equation that has two equal real roots. In this case, the system neither oscillates nor takes a longer time to return to its equilibrium position. The displacement formula for a critically damped system is: \[x(t) = (A + Bt)e^{-\omega_0 t}\] where A and B are constants determined by initial conditions, t is time, and ω₀ is the angular frequency.
02

Apply the initial velocity

The given initial velocity is -2 m/s. We can differentiate the displacement formula with respect to time to get the velocity formula: \[v(t) = \frac{d}{dt}x(t) = (B - A \omega_0t)e^{-\omega_0 t} - B\omega_0 te^{-\omega_0 t}\] At t=0, v(0)=-2, and in that case, we have: \[-2 = Be^{-\omega_0 \cdot 0} - 0 = B\] Thus, B = -2.
03

Determine the initial displacement

At t=0, x(0) = A + B0: \[x(0) = A + 0 = A\] The exercise does not provide the initial displacement value, so we will leave A as is and proceed to find the overshoot based on A.
04

Find the maximum displacement

To find the maximum displacement, we need to find the time when dx/dt, i.e., the velocity formula is equal to 0: \[v(t) = (B - A \omega_0t)e^{-\omega_0 t} - B\omega_0 te^{-\omega_0 t}=0\] This equation has two terms: the first part contains the unknowns, A and ω₀, whereas the second part only has B and ω₀. Using the initial condition B = -2, the equation becomes: \[(-2 - A \omega_0t)e^{-\omega_0 t} + 2\omega_0 te^{-\omega_0 t}=0\] We cannot solve this equation to get an exact value for the overshoot without any further information about A or ω₀. However, we can express the overshoot in terms of A and ω₀.
05

Express overshoot in terms of A and ω₀

The overshoot of the system is the maximum value of x(t) beyond the initial displacement (A): \[Overshoot = x_{max} - A\] By substituting the previously found expression for x(t) and B, we have: \[Overshoot = ((A - 2t)e^{-\omega_0t_{max}} - A)\] where t_max is the time at maximum displacement. With the given initial conditions and mathematical formulation of the critically damped system, we can only express the overshoot in terms of A, t, and ω₀. Further information about the system parameters is required to compute the exact value of the overshoot.

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Key Concepts

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

Differential Equations
At their core, differential equations are mathematical tools that describe the relationship between a function and its derivatives. In the context of physical systems, they often model the change in a system's state over time. A critically damped system, which we're interested in, can be represented by a second-order differential equation. This type of equation typically includes a term with the second derivative of the function (acceleration), a first derivative term (velocity), and a term with the function itself (displacement).

For a critically damped system, the equation has a specific form, which results in two equal real roots when solved. The general solution to the critically damped equation, as provided, is
\[x(t) = (A + Bt)e^{-\text{\(\omega_0\)}}t\],
where A and B are constants determined by initial conditions, t is time, and \(\omega_0\) is the system's angular frequency. Understanding this differential equation is crucial for predicting the behavior of a system that's been disturbed from its rest position.
Initial Conditions
Initial conditions play a crucial role in determining the specific solution to a differential equation that corresponds to a physical situation. They are essentially the 'starting values' for the state variables of the system at t = 0. For the problem at hand, we have a given initial velocity \(v_0 = -2 \text{ m/s}\).

These initial conditions allow us to determine the constants in the general solution of the differential equation. In the critically damped system's case, once the initial velocity is set, we can deduce one of the constants (B) by differentiating the displacement function and setting t = 0. However, without the initial displacement, the constant A remains undefined, illustrating how initial conditions can be partial and still provide some information about the system's future behavior but not complete the picture without additional data.
Overshoot Calculation
Overshoot refers to the extent to which a system exceeds its final, steady-state value after being disturbed from equilibrium. Calculating overshoot is important in understanding system performance, particularly in control systems where precision is key. In a critically damped system, ideally, there should be no overshoot because the system doesn't oscillate and should reach its equilibrium position without surpassing it.

In practice, to calculate overshoot, one would find the time at which the first derivative of the displacement (the velocity) is zero, indicating that the system has reached a peak or trough. For a critically damped system, this involves solving the differential equation with respect to time when \(v(t) = 0\). However, as seen in the step-by-step solution, without complete initial conditions - specifically, the initial displacement and system parameters like \(\omega_0\) - we can't definitively calculate the overshoot value. Instead, the overshoot is expressed in terms of the remaining unknowns, showcasing the system's behavior qualitatively rather than quantitatively.

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Most popular questions from this chapter

Two packages are placed on a spring scale whose plate weighs \(10 \mathrm{lb}\) and whose stiffness is \(50 \mathrm{lb} / \mathrm{in}\). When one package is accidentally knocked off the scale the remaining package is observed to oscillate through 3 cycles per second. What is the weight of the remaining package?

The cranking device shown consists of a mass-spring system of stiffiness \(k\) and mass \(m\) that is pin-connected to a massless rod which, in turn, is pin- connected to a wheel at radius \(R\), as indicated. If the mass moment of inertia of the wheel about an axis through the hub is \(I_{O}\), determine the natural frequency of the system. (The spring is unstretched when connecting pin is directly over hub ' \(O\) '.)

Determine the overshoot of the system of Problem \(2.25\) if it is critically damped and \(v_{0}=-4 \mathrm{~m} / \mathrm{sec}\).

A single degree of freedom system is represented as a \(4 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(6 \mathrm{~N} / \mathrm{m}\). If the coefficients of static and kinetic friction between the mass and the surface it moves on are \(\mu_{s}=\mu_{k}=0.1\), and the mass is displaced 2 meters to the right and released with a velocity of \(4 \mathrm{~m} / \mathrm{sec}\), determine the time after release at which the mass sticks and the corresponding displacement of the mass.

A single degree of freedom system is represented as a \(2 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(4 \mathrm{~N} / \mathrm{m}\). If the coefficients of static and kinetic friction between the block and the surface it moves on are respectively \(\mu_{s}=0.12\) and \(\mu_{k}=0.10\), determine the drop in amplitude between successive periods during free vibration. What is the frequency of the oscillations?

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