/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 64 Four passengers with combined ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 64

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

Short Answer

Expert verified
The period of vibration of the empty car, computed using the mass of the empty car and the spring constant, gives the final answer.

Step by step solution

01

Calculate the Spring Constant

First, calculate the spring constant using Hooke's Law. The equation is \(F = -kx\), where \(F\) is the force, \(k\) is the spring constant, and \(x\) is the displacement. Here the force \(F\) due to the added passengers is their weight which is their mass times gravity (\(F = mg\)), with \(m = 250 \mathrm{~kg}\) and \(g = 9.8 \mathrm{~m/s^2}\), and the displacement \(x = 0.04 \mathrm{~m}\). So, the spring constant \(k = -F/x\).
02

Calculate the Mass of Car Along with Passengers

Work out the total mass of the car together with the passengers by using the formula for the period of oscillations, rearranging it to solve for \(m\). The formula is \(T = 2\pi (m/k)^{0.5}\), with \(T = 1.92 \mathrm{~s}\), \(k\) from the previous step, and \(m\) being the mass we want to find. Hence, \(m = (T/2\pi)^2 * k \).
03

Determine the Mass of the Empty Car

Find the mass of the empty car by subtracting the mass of the passengers from the total mass found in step 2. As the mass of passengers is given as 250 kg, the mass of the empty car equals \(m_{\text{total}} - 250 \mathrm{~kg}\).
04

Compute the Period of the Empty Car

Calculate the period of vibration of the empty car \(T'\) using the formula from step 2, but this time with the mass of the empty car \(m_{\text{car}}\) and the same spring constant \(k\). Thus, \(T' = 2\pi (m_{\text{car}}/k)^{0.5}\).

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 91影视!

Key Concepts

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

Hooke's Law
Understanding Hooke's Law is fundamental to analyzing the behavior of springs and elastic materials. Simply put, Hooke's Law states that the force needed to extend or compress a spring by some distance is proportional to that distance.

The law is mathematically expressed as \( F = -kx \), where \( F \) represents the force applied on the spring, \( x \) is the distance the spring is stretched or compressed from its original equilibrium position, and \( k \) is a constant characteristic of the spring, known as the spring constant. The negative sign indicates that the force exerted by the spring is in the opposite direction to the displacement.

Let's take an example from our exercise where the car's springs compress due to the weight of the passengers. If the springs are compressed by \( 4.00 \text{cm} \) when the passengers enter the car, then the force exerted by the passengers (their weight) can be set equal to the spring force to find the spring constant \( k \). This relationship is pivotal for determining other dynamic properties of the spring-mass system such as its period of vibration.
Spring Constant
The spring constant, denoted by \( k \), is a measure of a spring's stiffness. It plays a critical role in the dynamics of spring-mass systems. In the context of Hooke's Law, the spring constant relates the force exerted by the spring to the displacement caused in that spring. Specifically, the higher the spring constant, the stiffer the spring, and thus the more force required to compress or extend it by a certain amount.

In our problem involving the car, the spring constant determines how the car will oscillate when it's set in motion, and is calculated through the equation derived from Hooke's Law. By knowing the weight of the passengers and the displacement of the springs, we can determine the spring constant. This constant remains the same regardless of whether the car is empty or loaded, stipulating that the spring's intrinsic property does not change with different loads.
Oscillation
Oscillation occurs when an object experiences a repetitive back-and-forth movement around a central point, or equilibrium position. In the case of the car and its passengers, when the passengers sit inside the car and compress the springs, and then when they depart, allowing the springs to extend back out, a cycle of oscillations could commence.

The period of vibration, which is the time for one complete cycle of oscillation, is of particular interest as it describes the oscillatory motion's tempo. The formula for the period \(T\) of a mass-spring system is given by \( T = 2\pi\sqrt{m/k} \), where \( m \) is the mass of the object attached to the spring and \( k \) is the spring constant. From our exercise solution, we see that if we have the spring constant and the mass of the car (with or without the passengers), we can easily calculate the car's period of vibration.

Understanding these concepts is not only critical for solving physics problems but also for appreciating the principles that govern many mechanical systems in engineering and nature.

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 slender rod of length \(80.0 \mathrm{~cm}\) and mass \(0.400 \mathrm{~kg}\) has its center of gravity at its geometrical center. But its density is not uniform; it increases by the same amount from the center of the rod out to either end. You want to determine the moment of inertia \(I_{\mathrm{cm}}\) of the rod for an axis perpendicular to the rod at its center, but you don't know its density as a function of distance along the rod, so you can't use an integration method to calculate \(I_{\mathrm{cm}}\). Therefore, you make the following measurements: You suspend the rod about an axis that is a distance \(d\) (measured in meters) above the center of the rod and measure the period \(T\) (measured in seconds) for small-amplitude oscillations about the axis. You repeat this for several values of \(d\). When you plot your data as \(T^{2}-4 \pi^{2} d / g\) versus \(1 / d\), the data lie close to a straight line that has slope \(0.320 \mathrm{~m} \cdot \mathrm{s}^{2} .\) What is the value of \(I_{\mathrm{cm}}\) for the rod?

Two uniform solid spheres, each with mass \(M=0.800 \mathrm{~kg}\) and radius \(R=0.0800 \mathrm{~m},\) 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 \(k=160 \mathrm{~N} / \mathrm{m}\) has one end attached to the wall and the other end attached to a friction less 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.

An unhappy \(0.300 \mathrm{~kg}\) rodent, moving on the end of a spring with force constant \(k=2.50 \mathrm{~N} / \mathrm{m},\) is acted on by a damping force \(F_{x}=-b v_{x}\). (a) If the constant \(b\) has the value \(0.900 \mathrm{~kg} / \mathrm{s},\) what is the frequency of oscillation of the rodent? (b) For what value of the constant \(b\) will the motion be critically damped?

BIO "Seeing" Surfaces at the Nanoscale. One technique for making images of surfaces at the nanometer scale, including membranes and biomolecules, is dynamic atomic force microscopy. In this technique, a small tip is attached to a cantilever, which is a flexible, rectangular slab supported at one end, like a diving board. The cantilever vibrates, so the tip moves up and down in simple harmonic motion. In one operating mode, the resonant frequency for a cantilever with force constant \(k=1000 \mathrm{~N} / \mathrm{m}\) is \(100 \mathrm{kHz}\). As the oscillating tip is brought within a few nanometers of the surface of a sample (as shown in the figure), it experiences an attractive force from the surface. For an oscillation with a small amplitude (typically, \(0.050 \mathrm{nm}),\) the force \(F\) that the sample surface exerts on the tip varies linearly with the displacement \(x\) of the tip, \(|F|=k_{\text {surf }} x,\) where \(k_{\text {surf }}\) is the effective force constant for this force. The net force on the tip is therefore \(\left(k+k_{\text {surf }}\right) x\), and the frequency of the oscillation changes slightly due to the interaction with the surface. Measurements of the frequency as the tip moves over different parts of the sample's surface can provide information about the sample.

Consider the system of two blocks and a spring shown in Fig. \(\mathrm{P} 14.66 .\) The horizontal surface is friction less, but there is static friction between the two blocks. The spring has force constant \(k=150 \mathrm{~N} / \mathrm{m} .\) The masses of the two blocks are \(m=0.500 \mathrm{~kg}\) and \(M=4.00 \mathrm{~kg} .\) You set the blocks into motion by releasing block \(M\) with the spring stretched a distance \(d\) from equilibrium. You start with small values of \(d,\) and then repeat with successively larger values. For small values of \(d,\) the blocks move together in SHM. But for larger values of \(d\) the top block slips relative to the bottom block when the bottom block is released. (a) What is the period of the motion of the two blocks when \(d\) is small enough to have no slipping? (b) The largest value \(d\) can have and there be no slipping is \(d=8.8 \mathrm{~cm} .\) What is the coefficient of static friction \(\mu_{\mathrm{s}}\) between the surfaces of the two blocks?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Mechanics and Materials

Read Explanation

Wave Optics

Read Explanation

Classical Mechanics

Read Explanation

Scientific Method Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation
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.