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

Four passengers with combined mass 250 kg compress the springs of a car with worn-out shock absorbers by 4.00 cm when they get in. Model the car and passengers as a single body on a single ideal spring. If the loaded car has a period of vibration of 1.92 s, what is the period of vibration of the empty car?

Short Answer

Expert verified
The period of vibration of the empty car is 1.72 s.

Step by step solution

01

Understanding the System

Given that the car and passengers are modeled as a body on a spring, the spring constant can be determined using Hooke's Law. We know that the combined weight of passengers (250 kg) compresses the spring by 4.00 cm (0.04 m).
02

Calculate the Spring Constant

Use Hooke's Law given by \[ F = kx \]where \( F \) is the force due to weight, \( k \) is the spring constant, and \( x \) is the displacement. Here, \( F = mg = 250 \times 9.8 \) N, and \( x = 0.04 \) m. Solve for \( k \):\[ k = \frac{250 \times 9.8}{0.04} = 61250 \, \text{N/m} \]
03

Determine the Loaded Period

We are given that the loaded car (car + passengers) has a period of 1.92 s. The period \( T \) of a mass-spring system is given by:\[ T = 2\pi\sqrt{\frac{m}{k}} \]Using this, rearrange to find \( m \) (loaded mass):\[ 1.92 = 2\pi\sqrt{\frac{m}{61250}} \]Solve for \( m \).
04

Find Total Loaded Mass

Solve \( 1.92 = 2\pi\sqrt{\frac{m}{61250}} \).\[ \frac{1.92}{2\pi} = \sqrt{\frac{m}{61250}} \]\[ \left(\frac{1.92}{2\pi}\right)^2 \times 61250 = m \]The loaded mass \( m \) includes car plus passengers.
05

Calculate Empty Car Mass

As the passengers have a combined mass of 250 kg:\[ m_\text{car alone} = m - 250 \]
06

Determine Empty Car Period

Now, use the empty car mass to find its period \( T_\text{empty} \):\[ T_\text{empty} = 2\pi\sqrt{\frac{m_\text{car alone}}{61250}} \]Calculate for \( T_\text{empty} \).
07

Calculation Summary

Upon calculation, using the determined empty car mass, we find:\[ T_\text{empty} = 1.72 \, ext{s} \]

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.

Spring Constant
The spring constant is a measure of a spring's stiffness. It tells us how much force is needed to compress or stretch the spring by a unit length. When we think of a car with worn-out shock absorbers, the spring constant determines how much the car will lower when passengers get in.
In our exercise, we found the spring constant using the formula from Hooke’s Law:
  • Force (\( F = mg \)) involves weight, which is mass times gravitational acceleration.
  • The displacement (\( x \) is the compression of the spring.
This allowed us to calculate the spring constant (\( k \) as 61250 N/m, indicating the car's springs are quite stiff.
Hooke's Law
Hooke's Law explains the behavior of springs in linear terms. It shows that the displacement of the spring is directly proportional to the force applied. The law's formula is\[ F = kx \]where:
  • \( F \) is the force applied in Newtons.
  • \( k \) is the spring constant in N/m.
  • \( x \) is the displacement in meters.
In this context, when 250 kg of passengers enter the car, they cause a displacement of 0.04 m. This displacement is due to the force exerted by the passengers' weight, allowing us to use Hooke's Law to find how strong the spring is.
Period of Vibration
The period of vibration refers to how long it takes for one complete cycle of motion in a simple harmonic oscillator, like a mass-spring system. It depends on both the mass and the spring constant.
The formula to find the period (\( T \) is given by:\[ T = 2\pi\sqrt{\frac{m}{k}} \]We use this to calculate the period of the loaded car and then the empty car.
For our loaded car, the period was given as 1.92 seconds. This information helps calculate the total mass involved and later determine the empty car's period of vibration.
Mass-Spring System
The concept of a mass-spring system is a fundamental example of harmonic motion. In this system, the mass oscillates back and forth after being displaced from its equilibrium position. This setup is beautifully demonstrated in the car model with passengers and the spring acting as the shock absorber.
In our problem:
  • The mass is the combined weight of the car and the passengers.
  • The spring is modeled by the car's suspension.
By understanding how these components interact, we can analyze how the system's period changes when mass (passengers) is added or removed. This gives us a clear picture of how modifications, like adding weight, affect the dynamics of the mass-spring system.

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 10.0-kg mass is traveling to the right with a speed of 2.00 m/s on a smooth horizontal surface when it collides with and sticks to a second 10.0-kg mass that is initially at rest but is attached to a light spring with force constant 170.0 N/m. (a) Find the frequency, amplitude, and period of the subsequent oscillations. (b) How long does it take the system to return the first time to the position it had immediately after the collision?

A 40.0-N force stretches a vertical spring 0.250 m. (a) What mass must be suspended from the spring so that the system will oscillate with a period of 1.00 s? (b) If the amplitude of the motion is 0.050 m and the period is that specified in part (a), where is the object and in what direction is it moving 0.35 s after it has passed the equilibrium position, moving downward? (c) What force (magnitude and direction) does the spring exert on the object when it is 0.030 m below the equilibrium position, moving upward?

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it is displaced 0.120 m from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency.

A 50.0-g hard-boiled egg moves on the end of a spring with force constant \(k =\) 25.0 N/m. Its initial displacement is 0.300 m. A damping force \(F_x = -bv_x\) acts on the egg, and the amplitude of the motion decreases to 0.100 m in 5.00 s. Calculate the magnitude of the damping constant \(b\).

Two uniform solid spheres, each with mass \(M =\) 0.800 kg and radius \(R =\) 0.0800 m, are connected by a short, light rod that is along a diameter of each sphere and are at rest on a horizontal tabletop. A spring with force constant \(k =\) 160 N/m has one end attached to the wall and the other end attached to a frictionless ring that passes over the rod at the center of mass of the spheres, which is midway between the centers of the two spheres. The spheres are each pulled the same distance from the wall, stretching the spring, and released. There is sufficient friction between the tabletop and the spheres for the spheres to roll without slipping as they move back and forth on the end of the spring. Show that the motion of the center of mass of the spheres is simple harmonic and calculate the period.

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