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(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?

Short Answer

Expert verified
(a) Position: \( x(t) = v_o t e^{-\gamma t} \). (b) Final distance: \( -x_o e^{-2\pi} \).

Step by step solution

01

Understand the Problem Context

In this exercise, we need to determine the position of a critically damped harmonic oscillator under two different initial conditions. A critically damped oscillator returns to equilibrium as quickly as possible without oscillating. Mathematically, this system is represented by a second-order differential equation of the form \( m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \) where \( m \) is mass, \( c \) is damping coefficient, and \( k \) is spring constant.
02

Initial Conditions: Part (a)

For part (a), the cart starts at equilibrium with an initial velocity \( v_o \). The equation for a critically damped system with these initial conditions is: \[ x(t) = (A + Bt) e^{-\gamma t} \] where \( \gamma = \frac{c}{2m} \), \( A \) and \( B \) are determined by initial conditions. Since \( x(0) = 0 \) and \( \frac{dx}{dt}|_{t=0} = v_o \), we can set \( A = 0 \) and we solve \( B = v_o + \gamma A \) to find \( B = v_o \). Therefore, \[ x(t) = v_ot e^{-\gamma t}. \]
03

Sketch of Part (a) Solution

The graph of \( x(t) = v_ot e^{-\gamma t} \) starts at zero and initially increases since \( v_o > 0 \). It then approaches zero as \( t \to \infty \), without oscillating, due to the exponential term. The curve is concave down attributable to \( e^{-\gamma t} \).
04

Initial Conditions: Part (b)

For part (b), the cart is released from rest at position \( x_o \). The system's equation is similarly: \[ x(t) = (A + Bt) e^{-\gamma t} \]. With initial conditions \( x(0) = x_o \) and \( \frac{dx}{dt}|_{t=0} = 0 \), we find \( A = x_o \) and \( B = -\gamma x_o \). Thus, \[ x(t) = (x_o - \gamma x_o t) e^{-\gamma t}. \]
05

Solve for Part (b) at Specific Time

To find the position at time \( \tau_o = \frac{2\pi}{\omega_o} \), substitute \( t = \tau_o \) into \[ x(t) \]: \\[ x(\tau_o) = (x_o - \gamma x_o \cdot \tau_o) e^{-\gamma \tau_o}. \] Since \( \gamma = \omega_o \) for critical damping in terms of natural frequency, \[ x(\tau_o) = x_o (1 - \omega_o \cdot \frac{2\pi}{\omega_o}) e^{-2\pi}. \] Simplifying, it reduces to \[ x(\tau_o) = - x_o e^{-2\pi}. \] Thus, the position is negative, indicating it's very close to equilibrium.

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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 equation that relates some function with its derivatives. In the context of a critically damped harmonic oscillator, the differential equation describes how the position of an object, like a cart on a spring, changes over time. For a system at critical damping, the equation is given by:
  • \( m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0 \)
Here, \( m \) is the mass, \( c \) the damping coefficient, and \( k \) the spring constant. This second-order linear differential equation captures the system’s dynamics. It shows how the forces acting on the object—namely the spring force, damping force, and inertia—balance out. In simpler terms, it describes the motion of the cart as it attempts to return to its equilibrium position. By solving this equation, we can predict how the system behaves under different initial conditions, whether it's kicked or released from a position.
Critical Damping
Critical damping is one of the key concepts when dealing with oscillating systems. It's the precise amount of damping that prevents oscillation and ensures the system returns to equilibrium as quickly as possible. This is different from underdamping, which allows the system to oscillate, and overdamping, which causes the system to return to equilibrium slowly. In a critically damped system, the damping coefficient \( c \) is related to the natural frequency \( \omega_o \) by the equation:
  • \( c = 2m\omega_o \)
Where \( m \) is the mass of the system. Critical damping is often used in applications such as car shock absorbers, where we wish to bring the car back to a stable position smoothly after encountering a bump. This type of damping is ideal in scenarios where quick settling without oscillation is necessary.
Initial Conditions
In the context of our problem, initial conditions are the known values at time \( t=0 \) which help us determine the constants in our solution. We deal with two cases for the critically damped oscillator:
  • Case (a): The cart starts with an initial velocity \( v_o \) and is at equilibrium, so \( x(0) = 0 \) and \( \frac{dx}{dt}|_{t=0} = v_o \).
  • Case (b): The cart is released from rest at position \( x_o \), so \( x(0) = x_o \) and \( \frac{dx}{dt}|_{t=0} = 0 \).
Initial conditions allow us to solve for the specific form of the solution, finding constants like \( A \) and \( B \) in the solution equations. For example, in case (a), the initial conditions simplify to \( x(t) = v_ot e^{-\gamma t} \), where \( \gamma \) is \( \frac{c}{2m} \). Initial conditions thus play a crucial role in defining the unique behavior of the system.
Harmonic Oscillator
A harmonic oscillator is a system that, when displaced from its equilibrium position, experiences a restoring force proportional to the displacement. For instance, springs and pendulums are classic examples. In our exercise, the cart attached to a spring is the harmonic oscillator.The behavior of this oscillator without damping (purely driven by the spring force) follows the equation:
  • \( m \frac{d^2x}{dt^2} + kx = 0 \)
However, our example focuses on a critically damped harmonic oscillator where the damping force is present to stop oscillations. "Harmonic" in this context refers to periodic, sinusoidal oscillations, which is what occurs when undamped. With critical damping, the oscillator returns to equilibrium without overshooting, achieving a state of balance as quickly as possible. This characteristic makes critically damped systems ideal in certain mechanical and electronic applications where rapid stabilization is desired.

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Most popular questions from this chapter

Let \(f(t)\) be a periodic function with period \(\tau\). Explain clearly why the average of \(f\) over one period is not necessarily the same as the average over some other time interval. Explain why, on the other hand, the average over a long time \(T\) approaches the average over one period, as \(T \rightarrow \infty\)

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