A shock absorber that provides critical damping with \(\omega_{\gamma}=72.4 \mathrm{~Hz}\) is compressed by \(6.41 \mathrm{~cm} .\) How far from the equilibrium position is it after \(0.0247 \mathrm{~s} ?\)

Short Answer

Expert verified
In a critically damped shock absorber with an initial displacement of 6.41 cm and damping angular frequency of 72.4 Hz, the displacement after 0.0247 seconds is approximately 6.05 cm away from its equilibrium position.

Step by step solution

01

Understand the concept of critical damping

In a critically damped or overdamped system, the system returns to its equilibrium position without oscillating. The damping is strong enough to prevent oscillations, but not too strong that it causes the return to equilibrium to be extremely slow.
02

Write down the known variables

From the problem statement, we know: \(\omega_{\gamma} = 72.4 \mathrm{~Hz}\) (damping angular frequency), Initial displacement \(x_0 = 6.41 \mathrm{~cm} = 0.0641 \mathrm{~m}\), (converted to meters) Time elapsed \(t = 0.0247 \mathrm{~s}\).
03

Find the solution to the critically damped oscillator equation

The equation for the displacement of a critically damped oscillator is given by: \(x(t) = (x_0 + ct)e^{-\omega_{\gamma}t}\), where x(t) is the displacement after time `t` and `c` is an arbitrary constant determined by the initial conditions of the system. In our case, since the shock absorber is initially compressed, we assume the initial velocity to be \(0\). So when \(t = 0\), \(v = 0\). From the equation of displacement, \(v(t) = \frac{dx(t)}{dt} = (-x_0 \omega_{\gamma} - c\omega_{\gamma})e^{-\omega_{\gamma}t}\). When \(t = 0\), \(v(0) = 0 = (-x_0 \omega_{\gamma} - c\omega_{\gamma})e^{0}\).
04

Calculate the constant 'c'

From the condition \(v(0)=0\), we have: \(0 = -x_0 \omega_{\gamma} - c\omega_{\gamma}\). Rearranging, we find the constant c: \(c = -x_0 = -0.0641 \mathrm{~m}\).
05

Calculate the displacement after the given time

Now that we have the expression for \(x(t)\) and the value of the constant 'c', we can find the displacement after time \(0.0247 \mathrm{~s}\): \(x(t) = (x_0 + ct)e^{-\omega_{\gamma}t}\). Plugging in the values, we get: \(x(0.0247) = (0.0641 - 0.0641)e^{-72.4*0.0247} \approx 0.0605 \mathrm{~m}\).
06

State the final answer

The shock absorber will be approximately \(6.05 \mathrm{~cm}\) away from its equilibrium position after \(0.0247 \mathrm{~s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Damped Oscillations
When systems capable of oscillating move back and forth around an equilibrium point, we observe what are known as oscillations. In the physical world, these oscillations don't go on forever; they gradually decrease in amplitude due to resistive forces—like friction or air resistance—which dissipate the system's energy. This process is called 'damping,' and the motion it results in is referred to as 'damped oscillations'.

Damped oscillations occur in various real-world systems, with one common example being the shock absorber in a vehicle. When pushed, the shock absorber moves back towards its original position but doesn't continue to move past it in a repetitive manner—as it would in simple harmonic motion—due to the damping force. Critical damping is a specific type of damped oscillation where the system returns to equilibrium in the shortest time without overshooting its original position. This behavior is particularly important in engineering and design, as it ensures stability and quick settling to rest, like in the case of automotive shock absorbers or in the prevention of excessive vibrations in tall buildings during strong winds or earthquakes.

To optimize understanding, visual aids such as graphs of displacement versus time for over-damped, under-damped, and critically damped systems can be very helpful. They demonstrate how the amplitude of oscillation evolves over time, helping students to visualize how critically damped systems differ from others.
Equilibrium Position
The equilibrium position in the context of oscillations is the point at which the net force acting on the system is zero. It is the position the system naturally assumes when there are no disturbances. This applies to a physical spring at rest, a pendulum hanging straight down, or, as in our exercise, a shock absorber that is neither compressed nor extended.

In the example of the shock absorber, the equilibrium position is particularly important because it's the state where the shock absorber isn't exerting any additional force to come back to its rest state. When discussing damped oscillations, understanding equilibrium is crucial as it is the final position the system aims to reach and stay at once the motion ceases. Regardless of the initial displacement or the type of damping involved, all damped oscillatory systems, in theory, aim to return to this stable point. By illustrating the concept of equilibrium with simple diagrams of forces at play, students can grasp the concept more effectively. Understanding equilibrium positions is fundamental to studying more advanced topics in mechanics and dynamics, as it forms the basis for analyzing system behavior.
Damping Angular Frequency
The damping angular frequency, often represented by the Greek letter omega with a subscript gamma \(\omega_{\gamma}\), is an important parameter in the study of damped oscillations. It is defined as the frequency at which the system would oscillate if there were no damping forces present—like the natural frequency—but adjusted for the effects of the damping. It essentially quantifies how fast the system returns to equilibrium without oscillating when there is critical damping.

Understanding damping angular frequency can be made simpler by connecting it to recognizable experiences. For example, consider striking a tuning fork and hearing its clear tone diminish over time due to damping. The rate at which it loses energy and quiets down involves the damping angular frequency. In the context of our exercise, with a damping angular frequency of 72.4 Hz, it dictates how rapidly the critically damped shock absorber reaches its equilibrium position after being disturbed. Calculations involving \(\omega_{\gamma}\) allow students to predict the future behavior of the system at specific time intervals after being displaced, as shown in the solution steps for the textbook problem. To assist learners further, practical demonstrations or simulations comparing different values of \(\omega_{\gamma}\) can vividly illustrate the effects of various damping scenarios on the time taken by the system to settle at the equilibrium position.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When the amplitude of the oscillation of a mass on a stretched string is increased, why doesn't the period of oscil lation also increase?

Two pendulums of identical length of \(1.000 \mathrm{~m}\) are suspended from the ceiling and begin swinging at the same time. One is at Manila, in the Philippines, where \(g=9.784 \mathrm{~m} / \mathrm{s}^{2}\) and the other is at Oslo, Norway, where \(g=9.819 \mathrm{~m} / \mathrm{s}^{2},\) After how many oscillations of the Manila pendulum will the two pendulums be in phase again? How long will it take for them to be in phase again?

A Foucault pendulum displayed in a museum is typically quite long, making the effect easier to see. Consider a Foucault pendulum of length \(15 \mathrm{~m}\) with a 110 -kg brass bob. It is set to swing with an amplitude of \(3.5^{\circ}\) a) What is the period of the pendulum? b) What is the maximum kinetic energy of the pendulum? c) What is the maximum speed of the pendulum?

A cylindrical can of diameter \(10.0 \mathrm{~cm}\) contains some ballast so that it floats vertically in water. The mass of can and ballast is \(800.0 \mathrm{~g}\), and the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\) The can is lifted \(1.00 \mathrm{~cm}\) from its equilibrium position and released at \(t=0 .\) Find its vertical displacement from equilibrium as a function of time. Determine the period of the motion. Ignore the damping effect due to the viscosity of the water.

With the right choice of parameters, a damped and driven physical pendulum can show chaotic motion, which is sensitively dependent on the initial conditions. Which statement about such a pendulum is true? a) Its long-term behavior can be predicted. b) Its long-term behavior is not predictable. c) Its long-term behavior is like that of a simple pendulum of equivalent length. d) Its long-term behavior is like that of a conical pendulum. e) None of the above is true.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free