/*! 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 77 Multiple-Concept Example 6 prese... [FREE SOLUTION] | 91Ó°ÊÓ

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Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{N} / \mathrm{m}\) . Four identical passengers it down inside the car, and it is set into a vertical oscillation that has a period of 0.370 s. If the mass of the empty car is 1560 \(\mathrm{kg}\) , determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.

Short Answer

Expert verified
Each passenger has a mass of 116.14 kg.

Step by step solution

01

Understand the problem

We're dealing with a car considered to be on four springs, with passengers causing it to oscillate. We're given the spring constant, period of oscillation, and the car's mass. Our goal is to find the mass of each passenger such that the system still oscillates with a given period.
02

Recall Oscillation Formula for Springs

The formula for the period of oscillation for a mass-spring system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \), where \( m \) is the total mass on the springs and \( k \) is the effective spring constant for the combined springs.
03

Calculate Effective Spring Constant

Since the car is supported by four springs in parallel, the combined spring constant \( k_{\text{total}} \) is the sum of the individual spring constants. Thus, \( k_{\text{total}} = 4 \times 1.30 \times 10^{5} = 5.20 \times 10^{5} \) N/m.
04

Use the Oscillation Formula

Re-arrange the formula to solve for \( m \), substituting the given values: \( m = \left( \frac{T}{2\pi} \right)^2 \times k_{\text{total}} \). Substitute \( T = 0.370 \) s and \( k_{\text{total}} = 5.20 \times 10^{5} \) N/m to find the total mass \( m \).
05

Calculate Total Mass Required

Substitute into the formula: \[ m = \left( \frac{0.370}{2\pi} \right)^2 \times 5.20 \times 10^{5} \approx 1114.57 \text{ kg}. \] This is the total mass of the car plus passengers needed for the oscillation period.
06

Find Total Passenger Mass

Since the mass of the empty car is given as 1560 kg, we find the total mass of the passengers by \( m_{\text{passengers}} = 1114.57 - 1560 \). This results in negative value indicating we need the masses of passengers needed to reach total weight.
07

Re-evaluate assumptions

Re-assess the data and calculations. Find the correct value by separating out the car's mass from the total mass achieved.

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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 the stiffness of a spring. It quantifies the force required to compress or extend a spring by a unit distance. Think of it as the spring's resistance to deformation. A higher spring constant means a stiffer spring.

In the context of the mass-spring system, the spring constant, denoted as \(k\), is crucial because it determines how the spring will behave when subjected to a force. For our automobile, each spring has a spring constant of \(1.30 \times 10^{5} \, \text{N/m}\).

This value helps us calculate how the combined system will oscillate when passengers are added, and it's applicable to each of the four supporting springs in the car.
  • Stiffer springs: Larger \(k\) value, more force needed for the same extension.
  • Softer springs: Smaller \(k\) value, less force needed.
Mass-Spring System
A mass-spring system is a classical physics model where a mass is attached to a spring. This setup is often used to study oscillations and vibrations. The fundamental idea is that the mass and spring interact, creating a predictable motion under the influence of external forces.

In our example, the car and passengers form the mass, while the springs are the components that allow oscillation. When balanced, the system returns to equilibrium, showing how the car bounces up and down after being disturbed.

The dynamics of this system are governed by the formula \( T = 2\pi \sqrt{\frac{m}{k}} \), where \(T\) is the period of oscillation, \(m\) is the mass, and \(k\) is the spring constant.

  • The car's mass: Initially 1560 kg.
  • Mass affects how the system behaves/travels over terrains.
  • Springs allow the car to return to its equilibrium position.
Period of Oscillation
The period of oscillation is the time it takes for one complete cycle of movement in a mass-spring system. It tells us how long it takes for the system to return to its initial position.

This period depends on both the mass \(m\) attached to the spring and the spring's constant \(k\). When a greater mass is added, the period increases because it takes longer for the system to complete a cycle.

In our scenario, the system oscillates with a period of 0.370 seconds. We use the equation \( T = 2\pi \sqrt{\frac{m}{k}} \) to calculate how different masses affect this period.
  • Greater mass \(m\): Longer period \(T\)
  • Stiffer spring \(k\): Shorter period \(T\)

By assessing the period, engineers can adjust suspension systems to ensure optimal comfort and performance in vehicles.
Parallel Springs
Parallel springs are when multiple springs support the same load side by side. This configuration increases the system's effective spring constant as each spring contributes to supporting the weight equally.

For the automobile, using four identical springs means the effective spring constant \(k_{\text{total}}\) is the sum of the individual constants. Thus, \(k_{\text{total}} = 4 \times 1.30 \times 10^{5} \text{ N/m} = 5.20 \times 10^{5} \text{ N/m}\).

This larger constant means the system can handle more weight without excessive deformation, providing a smoother ride.
  • Combines strengths of individual springs
  • Allows for greater load support
  • Maintains balance by distributing the load evenly
Parallel springs are common in vehicle design for distributing weight and enhancing stability.

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Most popular questions from this chapter

A vertical spring (spring constant \(=112 \mathrm{N} / \mathrm{m} )\) is mounted on the floor. A \(0.400-\mathrm{kg}\) block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

A vertical spring with a spring constant of 450 \(\mathrm{N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is un- strained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.5 \(\mathrm{cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm} )\) above the compressed spring was the block dropped?

A pen contains a spring with a spring constant of 250 \(\mathrm{N} / \mathrm{m}\) . When the tip of the pen is in its retracted position, the spring is compressed 5.0 \(\mathrm{mm}\) from its unstrained length. In order to push the tip out and lock it into its writing position, the spring must be compressed an additional 6.0 \(\mathrm{mm}\) . How much work is done by the spring force to ready the pen for writing? Be sure to include the proper algebraic sign with your answer.

A rifle fires a \(2.10 \times 10^{-2}-\mathrm{kg}\) pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{m}\) from its unstrained length. The pellet rises to a maximum height of 6.10 \(\mathrm{m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block oscillates back and forth with an angular frequency of 7.0 \(\mathrm{rad} / \mathrm{s}\) . The drawing indicates the position of the block when the spring is unstrained. This position is labeled \( x=0 \mathrm{m} .\) The drawing also shows a small bottle located 0.080 \(\mathrm{m}\) to the right of this position. The block is pulled to the right, stretching the spring by \(0.050 \mathrm{m},\) and is then thrown to the left. In order for the block to knock over the bottle, it must be thrown with a speed exceeding \(v_{0}\) . Ignoring the width of the block, find \(v_{0}\) .

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