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Chapter 15: Problem 50

Four people, each with a mass of \(72.4 \mathrm{kg},\) are in a car with a mass of \(1130 \mathrm{kg} .\) An earthquake strikes. The driver manages to pull off the road and stop, as the vertical oscillations of the ground surface make the car bounce up and down on its suspension springs. When the frequency of the shaking is \(1.80 \mathrm{Hz}\), the car exhibits a maximum amplitude of vibration. The earthquake ends, and the four people leave the car as fast as they can. By what distance does the car's undamaged suspension lift the car body as the people get out?

Short Answer

Expert verified
The car's suspension lifts the car body by approximately 6.2 cm when the four people exit the car.

Step by step solution

01

Get Initial Total Mass

Firstly, calculate the force of gravity on the total mass which will be the weight of the car with passengers, hence total mass is \( m_{1} = M_{car} + 4 \times m_{human} = 1130kg + 4 \times 72.4kg = 1417.6kg.\)
02

Calculate Spring Constant

Knowing the frequency of oscillations when the car is fully loaded, use the formula for the period of a mass-spring system: \( T = 2\pi \sqrt{\frac{m}{k}} \), where \( T \) is the period, \( m \) is mass and \( k \) is the spring constant. Rearranging to find \( k \), we get : \( k = \frac{4\pi^{2}m}{{T^{2}}} = \frac{4\pi^{2}1417.6}{{(1/1.8)^{2}}} = 179878.8 N/m \)
03

Find mass of car and calculate change in height

Now without the people, the car has a mass \( m_{2} = 1130kg \). So, the net force acting on the car (ignoring friction and air resistance for now) is the force of gravity. The car will go up until this force is balanced by the spring force, as given by Hooke's Law: \( F_{spring} = k \times \Delta x \), where \( \Delta x \) is displacement. Equating \( mg = k \times \Delta x\), we solve for \( \Delta x = \frac{mg}{k} = \frac{1130 \times 9.8}{179878.8} = 0.062m = 6.2cm.\)

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Key Concepts

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

Understanding Hooke's Law
Hooke's Law is a fundamental principle in physics that relates the force needed to extend or compress a spring to the distance the spring is stretched or compressed. Mathematically, this law is expressed as:
\[ F = -kx \] where:
  • \( F \) is the force applied on the spring,
  • \( k \) is the spring constant, which measures the stiffness of the spring,
  • \( x \) is the displacement of the spring from its rest position.

The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement; this is what causes the spring to return to its equilibrium position after being disturbed. In the context of the problem, Hooke's Law helps us determine how much the car will rise (i.e., the displacement of the spring) when the weight of the passengers is removed.
Spring Constant Calculation Explained
The spring constant, denoted by \( k \) in Hooke's Law, is a measure of spring stiffness. A higher spring constant means a stiffer spring. To calculate the spring constant, you can use oscillation data from a system where the spring is part of a mass-spring setup undergoing simple harmonic motion. From the given problem, we calculate the spring constant using the formula:
\[ k = \frac{4\bar{\beta}^{2}m}{{T^{2}}} \] wherein \( T \) is the period of oscillation and \( m \) is the mass causing the spring to oscillate.
After rearranging this formula considering the frequency of oscillation (\( f = \frac{1}{T} \) ), we obtained the spring constant necessary to solve for the lift of the car body. This is critical to understanding how the system behaves under varying loads.
Simple Harmonic Motion Overview
Simple harmonic motion (SHM) refers to a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that displacement. It's characterized by its sine or cosine waveform when graphed over time, and it's the type of motion exhibited by mass-spring systems when they are not experiencing damping or external forces.

Characteristics of SHM

For a system in SHM, there are a few key characteristics:
  • The motion is oscillatory, moving back and forth over a central point.
  • It has a constant frequency and period that depend only on the mass and spring constant (for mass-spring systems).
  • Energy transformation between potential and kinetic occurs continuously.

In our car suspension example, the vehicle's bouncing on the springs forms an SHM system. The earthquake-induced oscillation frequency helps us determine the spring constant and eventually the car's lift in response to the passengers' exit.

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Most popular questions from this chapter

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the expression $$x=(5.00 \mathrm{cm}) \cos (2 t+\pi / 6)$$. where \(x\) is in centimeters and \(t\) is in seconds. At \(t=0\) find (a) the position of the piston, (b) its velocity, and (c) its acceleration. (d) Find the period and amplitude of the motion.

A \(2.00-\mathrm{kg}\) object attached to a spring moves without friction and is driven by an external force given by \(F=\) \((3.00 \mathrm{N}) \sin (2 \pi t) .\) If the force constant of the spring is 20.0 N/m, determine (a) the period and (b) the amplitude of the motion.

A weight of \(40.0 \mathrm{N}\) is suspended from a spring that has a force constant of \(200 \mathrm{N} / \mathrm{m} .\) The system is undamped and is subjected to a harmonic driving force of frequency \(10.0 \mathrm{Hz}\) resulting in a forced-motion amplitude of \(2.00 \mathrm{cm} .\) Determine the maximum value of the driving force.

A \(2.00-\mathrm{kg}\) object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of \(20.0 \mathrm{N}\) is required to hold the object at rest when it is pulled \(0.200 \mathrm{m}\) from its equilibrium position (the origin of the \(x\) axis). The object is now released from rest with an initial position of \(x_{i}=0.200 \mathrm{m},\) and it subsequently undergoes simple harmonic oscillations. Find (a) the force constant of the spring, (b) the frequency of the oscillations, and (c) the maximum speed of the object. Where does this maximum speed occur? (d) Find the maximum acceleration of the object. Where does it occur? (e) Find the total energy of the oscillating system. Find (f) the speed and (g) the acceleration of the object when its position is equal to one third of the maximum value.

A particle of mass \(4.00 \mathrm{kg}\) is attached to a spring with a force constant of \(100 \mathrm{N} / \mathrm{m}\). It is oscillating on a horizontal frictionless surface with an amplitude of \(2.00 \mathrm{m} .\) A \(6.00-\mathrm{kg}\) object is dropped vertically on top of the \(4.00-\mathrm{kg}\) object as it passes through its equilibrium point. The two objects stick together. (a) By how much does the amplitude of the vibrating system change as a result of the collision? (b) By how much does the period change? (c) By how much does the energy change? (d) Account for the change in energy.
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