Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Cars have shock absorbers to damp the oscillations that would otherwise occur when the springs that attach the wheels to the car's frame are compressed or stretched. Ideally, the shock absorbers provide critical damping. If the shock absorbers fail, they provide less damping, resulting in an underdamped motion. You can perform a simple test of your shock absorbers by pushing down on one corner of your car and then quickly releasing it If this results in an up-and- down oscillation of the car, you know that your shock absorbers need changing. The spring on each wheel of a car has a spring constant of \(4005 \mathrm{~N} / \mathrm{m}\), and the car has a mass of \(851 \mathrm{~kg}\), equally distributed over all four wheels. Its shock absorbers have gone bad and provide only \(60.7 \%\) of the damping they were initially designed to provide. What will the period of the underdamped oscillation of this car be if the pushing-down shock absorber test is performed?

Short Answer

Expert verified
Answer: The period of the underdamped oscillation is approximately 12.97 seconds.

Step by step solution

01

Get the effective spring constant for one wheel

Since the car's mass is equally distributed over all four wheels, we can calculate the effective spring constant for one wheel by dividing the car's total spring constant by 4. \(k_{eff} = \frac{4005 \mathrm{~N/m}}{4} = 1001.25 \mathrm{~N/m}\)
02

Calculate the critical damping coefficient

The critical damping coefficient can be calculated by using the following formula: \(c_{crit} = 2 \sqrt{m k_{eff}}\) Plug in the mass m = 851 kg and effective spring constant \(k_{eff} = 1001.25 \mathrm{~N/m}\): \(c_{crit} = 2 \sqrt{(851 \mathrm{~kg})(1001.25 \mathrm{~N/m})} = 1841.90 \mathrm{~N s/m}\)
03

Find the actual damping coefficient

As the shock absorbers provide only 60.7% of the damping, multiply the critical damping coefficient by this percentage to find the actual damping coefficient: \(c = 0.607 \cdot c_{crit} = 0.607 \cdot 1841.90 \mathrm{~N s/m} = 1117.62 \mathrm{~N s/m}\)
04

Calculate the angular frequency of the underdamped oscillation

The angular frequency of the underdamped oscillation can be calculated using the following formula: \(\omega_d = \sqrt{\frac{k_{eff}}{m} - \frac{c^2}{4m^2}}\) Plug in the effective spring constant \(k_{eff} = 1001.25 \mathrm{~N/m}\), mass m = 851 kg, and actual damping coefficient \(c = 1117.62 \mathrm{~N s/m}\): \(\omega_d = \sqrt{\frac{1001.25 \mathrm{~N/m}}{851 \mathrm{~kg}} - \frac{(1117.62 \mathrm{~N s/m})^2}{4(851 \mathrm{~kg})^2}} = 0.4847 \mathrm{~rad/s}\)
05

Find the period of the underdamped oscillation

Finally, now that we have the angular frequency, we can calculate the period of the underdamped oscillation using the following formula: \(T = \frac{2 \pi}{\omega_d}\) Plug in the angular frequency \(\omega_d = 0.4847 \mathrm{~rad/s}\): \(T = \frac{2 \pi}{0.4847 \mathrm{~rad/s}} = 12.97 \mathrm{~s}\) Thus, the period of the underdamped oscillation of this car will be approximately 12.97 seconds if the pushing-down shock absorber test is performed.

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.

Understanding Shock Absorbers
Shock absorbers are essential components in vehicles that control the oscillations occurring when a car goes over bumps or uneven surfaces. They work alongside the springs attached to the wheels to absorb and dissipate the energy. While springs help absorb kinetic energy from the road, shock absorbers help control and dampen the motion. When they work efficiently, they provide critical damping. Critical damping prevents prolonged oscillations, leading to a stable and comfortable ride.
  • Prevents Excessive Motion: By slowing down the spring's oscillations.
  • Ensures Safety: By keeping tires in contact with the road, improving handling and braking.
If your car bounces excessively after being pushed, the shock absorbers might be underperforming. This means they function as underdamped systems, failing to offer the necessary level of damping.
The Role of Critical Damping
Critical damping is the ideal state where oscillations cease as quickly as possible without oscillating. It's a key concept in understanding shock absorbers, as it ensures minimal oscillation. The damping level in your car determines how quickly the motion is brought to rest without oscillating back and forth unnecessarily. Critical damping offers the perfect balance, halting motion in the shortest time possible.
  • Quick Stabilization: Returns the system to rest without bouncing.
  • Optimized Comfort: Provides a smooth ride by minimizing vibrating motion after a disturbance.
In the case of a car, properly functioning shock absorbers aim to achieve this condition. However, when shock absorbers are not functioning at full capacity, they only offer partial damping, leading to undesirable car dynamics.
Spring Constant and Its Importance
The spring constant, denoted as "k," measures a spring's stiffness. It defines how much force is needed to compress or stretch the spring by a specific amount. For vehicles, the spring constant determines how the car responds to weight distribution and road irregularities. It's crucial for the shock absorbers to have a correctly matched spring constant to maintain comfort and stability.
  • Represents Stiffness: A higher value indicates a stiffer spring.
  • Dictates Oscillation Rate: Affects how quickly the system oscillates.
In our example, with a spring constant of 4005 N/m for the entire car, each wheel's effective spring constant is 1001.25 N/m. This value is crucial for calculating other parameters like damping and angular frequency.
Exploring Angular Frequency in Oscillations
Angular frequency, represented by the symbol \( \omega \), is a measure of how quickly an object oscillates.In physics, it helps us describe the motion of oscillating systems like those involving shock absorbers and springs. It is particularly relevant when detailing underdamped systems.Angular frequency depends on several factors:
  • Spring Constant (k): Higher spring constants lead to higher angular frequencies.
  • Mass (m): Heavier systems oscillate slower, reducing the angular frequency.
  • Damping Coefficient (c): It also integrates the damping effect, crucial for underdamped systems.
For our car exercise, the angular frequency \( \omega_d \) of 0.4847 rad/s tells us how quickly the car will oscillate once the spring is compressed and released. This value is imperative for determining the period of oscillation and understanding vehicle dynamics.

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

A grandfather clock uses a pendulum and a weight. The pendulum has a period of \(2.00 \mathrm{~s}\), and the mass of the bob is 250. \(\mathrm{g}\). The weight slowly falls, providing the energy to overcome the damping of the pendulum due to friction. The weight has a mass of \(1.00 \mathrm{~kg}\), and it moves down \(25.0 \mathrm{~cm}\) every day. Find \(Q\) for this clock. Assume that the amplitude of the oscillation of the pendulum is \(10.0^{\circ}\)

A mass \(M=0.460 \mathrm{~kg}\) moves with an initial speed \(v=3.20 \mathrm{~m} / \mathrm{s}\) on a level frictionless air track. The mass is initially a distance \(D=0.250 \mathrm{~m}\) away from a spring with \(k=\) \(840 \mathrm{~N} / \mathrm{m}\), which is mounted rigidly at one end of the air track. The mass compresses the spring a maximum distance \(d\), before reversing direction. After bouncing off the spring the mass travels with the same speed \(v\), but in the opposite dircction. a) Determine the maximum distance that the spring is compressed. b) Find the total elapsed time until the mass returns to its starting point. (Hint: The mass undergoes a partial cycle of simple harmonic motion while in contact with the spring.)

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 \(0.0247 \mathrm{~s} ?\)

In a lab, a student measures the unstretched length of a spring as \(11.2 \mathrm{~cm}\). When a 100.0 - g mass is hung from the spring, its length is \(20.7 \mathrm{~cm}\). The mass-spring system is set into oscillatory motion, and the student obscrves that the amplitude of the oscillation decreases by about a factor of 2 after five complete cycles. a) Calculate the period of oscillation for this system, assuming no damping. b) If the student can measure the period to the nearest \(0.05 \mathrm{~s}\). will she be able to detect the difference between the period with no damping and the period with damping?

A 2.0 -kg mass attached to a spring is displaced \(8.0 \mathrm{~cm}\) from the equilibrium position. It is released and then oscillates with a frequency of \(4.0 \mathrm{~Hz}\) a) What is the energy of the motion when the mass passes through the equilibrium position? b) What is the speed of the mass when it is \(20 \mathrm{~cm}\) from the equilibrium position?

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