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Consider an underdamped oscillator (such as a mass on the end of a spring) that is released from rest at position \(x_{\mathrm{o}}\) at time \(t=0 .\) (a) Find the position \(x(t)\) at later times in the form $$x(t)=e^{-\beta t}\left[B_{1} \cos \left(\omega_{1} t\right)+B_{2} \sin \left(\omega_{1} t\right)\right]$$ That is, find \(B_{1}\) and \(B_{2}\) in terms of \(x_{\mathrm{o}}\). (b) Now show that if you let \(\beta\) approach the critical value \(\omega_{\mathrm{o}}\) your solution automatically yields the critical solution. (c) Using appropriate graphing software, plot the solution for \(0 \leq t \leq 20,\) with \(x_{\mathrm{o}}=1, \omega_{\mathrm{o}}=1,\) and \(\beta=0,0.02,0.1,0.3,\) and 1

Short Answer

Expert verified
(a) \( B_1 = x_0, B_2 = \frac{\beta x_0}{\omega_1} \). (b) Critical solution shows as \( \beta \to \omega_o \). (c) Plot shows damping effects.

Step by step solution

01

Understanding the Differential Equation

The system can be described by the second-order differential equation \( \frac{d^2 x}{dt^2} + 2\beta \frac{dx}{dt} + \omega_o^2 x = 0 \). For an underdamped oscillator, \( \beta < \omega_o \), and the solution takes the form \( x(t) = e^{-\beta t} (B_1 \cos(\omega_1 t) + B_2 \sin(\omega_1 t)) \), where \( \omega_1 = \sqrt{\omega_o^2 - \beta^2} \).
02

Apply Initial Conditions

Given that the system is released from rest at position \( x_0 \), this implies \( x(0) = x_0 \) and \( \frac{dx}{dt}(0) = 0 \). First, substitute \( t = 0 \) into the position equation to find \( B_1 \): \( x(0) = e^{0} (B_1 \cdot 1 + B_2 \cdot 0) = B_1 = x_0 \).
03

Differentiate and Apply Rest Condition

Differentiate \( x(t) \) with respect to \( t \) to find the velocity: \( \frac{dx}{dt} = -\beta e^{-\beta t}(B_1 \cos(\omega_1 t) + B_2 \sin(\omega_1 t)) + e^{-\beta t}(-B_1 \omega_1 \sin(\omega_1 t) + B_2 \omega_1 \cos(\omega_1 t)) \). At \( t = 0 \), this simplifies to \( -\beta x_0 + B_2 \omega_1 = 0 \). Solve for \( B_2 \): \( B_2 = \frac{\beta x_0}{\omega_1} \).
04

Confirm the Critical Solution

As \( \beta \to \omega_o \), the system approaches critical damping. Here, \( \omega_1 \to 0 \), and the behavior changes to a critically damped oscillator. In the critically damped case, the solution is often \( x(t) = (A + Bt)e^{-\omega_o t} \), which can be related by limits to the form \( e^{-\beta t}(B_1 \cos(\omega_1 t) + B_2 \sin(\omega_1 t)) \) as \( \omega_1 \to 0 \).
05

Graphing the Solution

Plot \( x(t) = e^{-\beta t}(B_1 \cos(\omega_1 t) + B_2 \sin(\omega_1 t)) \) for different values of \( \beta \) using software for the given conditions: \( x_0 = 1 \), \( \omega_o = 1 \), and the specified \( \beta \) values. Each plot shows how the damping affects the amplitude and oscillatory behavior over the range \( 0 \leq t \leq 20 \).

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Key Concepts

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

Differential Equation
A differential equation is a mathematical expression that relates a function to its derivatives. In the context of an underdamped oscillator, the system is governed by a second-order differential equation:
  • \( \frac{d^2 x}{dt^2} + 2\beta \frac{dx}{dt} + \omega_o^2 x = 0 \)
This equation describes how the position \( x(t) \) of the oscillator evolves over time under the influence of damping, represented by \( \beta \), and the natural frequency, \( \omega_o \). The equation is called 'second-order' because it involves second derivatives.
The solution to this differential equation helps predict the behavior of the oscillator, specifying how it moves over time. It's crucial for modeling systems where damping causes the system to lose energy, leading to a decrease in amplitude over time.
Understanding differential equations allows us to make predictions about physical systems, offering insights into phenomena like oscillation and damping.
Initial Conditions
Initial conditions in differential equations provide the specific values needed to find a particular solution from a family of solutions. For the underdamped oscillator, the initial conditions are:
  • \( x(0) = x_0 \), the position of the oscillator at time \( t = 0 \).
  • \( \frac{dx}{dt}(0) = 0 \), indicating the oscillator is released from rest.
These conditions allow us to determine the constants \( B_1 \) and \( B_2 \) in the solution. By inserting these initial conditions into our position and velocity equations, we can solve for these constants:
\( B_1 = x_0 \)
\( B_2 = \frac{\beta x_0}{\omega_1} \)
The initial conditions ensure that our mathematical model accurately reflects the physical scenario we're analyzing. They are pivotal for tailoring the general solution of the differential equation into one that fits the specific situation of the oscillator at the start of its motion.
Critical Damping
Critical damping refers to the specific condition where the system returns to equilibrium without oscillating and as quickly as possible. It occurs when the damping coefficient \( \beta \) reaches the critical value \( \omega_o \), the natural frequency:
  • When \( \beta = \omega_o \), we achieve critical damping.
  • The system does not oscillate but returns to its equilibrium position smoothly.
In a mathematical sense, as \( \beta \rightarrow \omega_o \), \( \omega_1 \rightarrow 0 \) which alters the form of the solution. The underdamped solution converges to a critically damped form:
\( x(t) = (A + Bt)e^{-\omega_o t} \)
This transformation highlights how the damping impacts the system’s behavior, shifting from oscillatory to a smooth return to equilibrium as damping increases. Critical damping is crucial in systems where a rapid return to stability is needed, such as in automotive suspensions or building structures to prevent excessive movement or damage.
Second-order Differential Equation
Second-order differential equations are fundamental in modeling physical systems involving acceleration, such as oscillators. They involve the second derivative of a function, indicating how the rate of change of the rate of change (acceleration) is described in terms of other variables:
  • Our equation \( \frac{d^2 x}{dt^2} + 2\beta \frac{dx}{dt} + \omega_o^2 x = 0 \) is second-order because it includes \( \frac{d^2 x}{dt^2} \).
In the context of the underdamped oscillator, this equation captures:
  • The natural motion described by \( \omega_o^2 x \).
  • The damping effect through \( 2\beta \frac{dx}{dt} \).
Second-order differential equations are widespread in engineering and physics for modeling all kinds of dynamical systems—from simple harmonic oscillators to complex mechanical systems. They allow for the precise description and prediction of system behaviors, making them invaluable in design, analysis, and control processes. Understanding these equations is key to unlocking how various systems react under different forces and constraints.

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Most popular questions from this chapter

Write down the potential energy \(U(\phi)\) of a simple pendulum (mass \(m,\) length \(l\) ) in terms of the angle \(\phi\) between the pendulum and the vertical. (Choose the zero of \(U\) at the bottom.) Show that, for small angles, \(U\) has the Hooke's law form \(U(\phi)=\frac{1}{2} k \phi^{2},\) in terms of the coordinate \(\phi .\) What is \(k ?\)

We know that if the driving frequency \(\omega\) is varied, the maximum response \(\left(A^{2}\right)\) of a driven damped oscillator occurs at \(\omega \approx \omega_{\mathrm{o}}\) (if the natural frequency is \(\omega_{\mathrm{o}}\) and the damping constant \(\beta \ll\) \(\omega_{\mathrm{o}}\) ). Show that \(A^{2}\) is equal to half its maximum value when \(\omega \approx \omega_{\mathrm{o}} \pm \beta,\) so that the full width at half maximum is just \(2 \beta\). [Hint: Be careful with your approximations. For instance, it's fine to say \(\left.\omega+\omega_{\mathrm{o}} \approx 2 \omega_{\mathrm{o}}, \text { but you certainly mustn't say } \omega-\omega_{\mathrm{o}} \approx 0 .\right]\)

When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscillates is each axle assembly, comprising the axle and its two wheels.) Find the speed of my car at which this oscillation resonates, given the following information: (a) When four \(80-\mathrm{kg}\) men climb into my car, the body sinks by a couple of centimeters. Use this to estimate the spring constant \(k\) of each of the four springs. (b) If an axle assembly (axle plus two wheels) has total mass \(50 \mathrm{kg}\), what is the natural frequency of the assembly oscillating on its two springs? ( \(\mathbf{c}\) ) If the bumps on a road are \(80 \mathrm{cm}\) apart, at about what speed would these oscillations go into resonance?

(a) Consider a cart on a spring which is critically damped. At time \(t=0\), it is sitting at its equilibrium position and is kicked in the positive direction with velocity \(v_{\mathrm{o}} .\) Find its position \(x(t)\) for all subsequent times and sketch your answer. (b) Do the same for the case that it is released from rest at position \(x=x_{\mathrm{o}} .\) In this latter case, how far is the cart from equilibrium after a time equal to \(\tau_{\mathrm{o}}=2 \pi / \omega_{\mathrm{o}}\) the period in the absence of any damping?

Another interpretation of the \(Q\) of a resonance comes from the following: Consider the motion of a driven damped oscillator after any transients have died out, and suppose that it is being driven close to resonance, so you can set \(\omega=\omega_{\mathrm{o}}\). (a) Show that the oscillator's total energy (kinetic plus potential) is \(E=\frac{1}{2} m \omega^{2} A^{2} .\) (b) Show that the energy \(\Delta E_{\text {dis dissipated during one cycle by the damping force }}\) \(F_{\text {dmp }}\) is \(2 \pi m \beta \omega A^{2} .\) (Remember that the rate at which a force does work is \(F v .\) ) (c) Hence show that \(Q\) is \(2 \pi\) times the ratio \(E / \Delta E_{\text {dis: }}\)

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