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Chapter 14: Problem 66

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:\[ T_\text{empty} = 1.72 \, ext{s} \]

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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鈥檚 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.

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Most popular questions from this chapter

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of the motion is 0.090 m, it takes the block 2.70 s to travel from \(x =\) 0.090 m to \(x = -\)0.090 m. If the amplitude is doubled, to 0.180 m, how long does it take the block to travel (a) from \(x =\) 0.180 m to \(x = -\)0.180 m and (b) from \(x =\) 0.090 m to \(x = -\)0.090 m?

A building in San Francisco has light fixtures consisting of small \(2.35-\mathrm{kg}\) bulbs with shades hanging from the ceiling at the end of light, thin cords \(1.50 \mathrm{~m}\) long. If a minor earthquake occurs, how many swings per second will these fixtures make?

This procedure has been used to "weigh" astronauts in space: A 42.5-kg chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes 1.30 s to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes 2.54 s for one cycle. What is the mass of the astronaut?

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.
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