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\(\bullet\) \(\bullet\) Four passengers with a combined mass of 250 \(\mathrm{kg}\) compress the springs of a car with wom-out shock absorbers by 4.00 \(\mathrm{cm}\) when they enter it. Model the car and passengers as a single body on a single ideal spring. If the loaded car has a period of vibration of \(1.08 \mathrm{s},\) what is the period of vibration of the empty car?

Short Answer

Expert verified
Calculate the mass of the car and then substitute to find the period of the empty car using spring constant.

Step by step solution

01

Understand the Problem

We need to determine the period of vibration of the car without passengers (empty car) using given data including the mass of passengers and the period of vibration when the car is loaded.
02

Define Given Quantities

- Combined mass of passengers, \( m = 250 \text{ kg} \)- Spring compression due to passengers, \( x = 4.00 \text{ cm} = 0.04 \text{ m} \)- Period of loaded car, \( T' = 1.08 \text{ s} \)
03

Relation between Period and Spring Constant

The period \( T' \) for a simple harmonic oscillator is given by \( T' = 2\pi \sqrt{\frac{m+M}{k}} \), where \( m \) is the mass of the passengers, \( M \) is the mass of the car, and \( k \) is the spring constant.
04

Calculate the Spring Constant

Using Hooke's Law, \( F = kx \), and the fact that the force is the weight of the passengers \( F = mg \), we have:\[k = \frac{mg}{x} = \frac{250 \cdot 9.81}{0.04} = 61250 \, \text{N/m}\]
05

Express Period of Empty Car

The period \( T \) of the empty car is given by:\[T = 2\pi \sqrt{\frac{M}{k}}\]We don't have the mass \( M \) of the car directly, but we can find it from the loaded car period.
06

Use Loaded Car Period to Find Car Mass

From the loaded car:\[1.08 = 2\pi \sqrt{\frac{m+M}{k}} \\Rightarrow \left(\frac{1.08}{2\pi}\right)^2 = \frac{250 + M}{61250} \M = 61250 \left(\frac{1.08}{2\pi}\right)^2 - 250\]

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Key Concepts

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

Physics Problem Solving
Physics problems often require us to break down given information and employ mathematical relationships to find a solution. In scenarios involving harmonic motion, understanding key concepts and how they relate to each other is crucial.

The given problem involves calculating the period of a vibration for an empty car using the data from when it is loaded. This requires an understanding of how mass, force, and spring constants interact.

To tackle problems like these:
  • First, identify what is known from the problem – variables such as mass, compression, and period.
  • Second, understand the relationship between these variables. For oscillatory motion, formulas like the period of a harmonic oscillator are essential.
  • Third, use relevant equations to connect the given information to what is unknown.
By methodically approaching each piece of the problem, complex physical concepts become more manageable.
Spring Constant Calculation
The spring constant, a crucial component in the context of harmonic oscillators, defines how resistant a spring is to being compressed or stretched. It's denoted by the symbol \(k\) and measured in Newtons per meter (N/m).

To find the spring constant in this problem, we apply Hooke's Law, which is formulated as \( F = kx \). Here:
  • \(F\) represents the force applied to the spring, equating to the gravitational force in this context, \( F = mg \).
  • \(k\) is what we need to find.
  • \(x\) is the displacement the spring undergoes due to the force, 0.04 meters, in our case.
Therefore, our equation to find the spring constant becomes: \ \[ k = \frac{mg}{x} = \frac{250 \times 9.81}{0.04} = 61250 \, \text{N/m} \]

The calculation provides a clear picture of the spring's stiffness, indicating how much force is needed for a given compression or stretch.
Simple Harmonic Motion
Simple harmonic motion (SHM) describes a type of oscillatory motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. Objects like pendulums or oscillating springs can exhibit this motion.

The period \( T \) of a simple harmonic oscillator such as the spring in our problem is given by the formula: \ \[ T = 2\pi \sqrt{\frac{M}{k}} \]where:
  • \(T\) is the period.
  • \(M\) is the system's mass (in our case, the car's mass).
  • \(k\) is the spring constant.
In our scenario, while the period of the loaded car is already known, finding the period of the empty car involves recalculating using the known spring constant and the car's calculated mass minus the passenger's weight.

Thus, understanding SHM principles allows us to predict and calculate motions of oscillatory systems.

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Most popular questions from this chapter

\(\bullet\) \(\bullet \mathrm{A} 100 \mathrm{kg}\) mass suspended from a wire whose unstretched length is 4.00 \(\mathrm{m}\) is found to stretch the wire by 6.0 \(\mathrm{mm}\) . The wire has a uniform cross-sectional area of 0.10 \(\mathrm{cm}^{2} .\) (a) If the load is pulled down a small additional distance and released, find the frequency at which it vibrates. (b) Compute Young's modulus for the wire.

\(\bullet\) A 0.500 kg glider on an air track is attached to the end of an ideal spring with force constant \(450 \mathrm{N} / \mathrm{m} ;\) it undergoes simple harmonic motion with an amplitude of 0.040 \(\mathrm{m} .\) Compute (a) the maximum speed of the glider, (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m},(\mathrm{c})\) the magnitude of the maximum acceleration of the glider, (d) the acceleration of the glider at \(x=-0.015 \mathrm{m},\) and \((\mathrm{e})\) the total mechanical energy of the glider at any point in its motion.

\(\bullet\) \(\bullet\) An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) 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)?

\(\bullet\) A 2.00 kg frictionless block is attached to an ideal spring with force constant 315 \(\mathrm{N} / \mathrm{m}\) . Initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 12.0 \(\mathrm{m} / \mathrm{s} .\) Find (a) the amplitude of the motion, (b) the maximum acceleration of the block, and (c) the maximum force the spring exerts on the block.

\(\bullet\) A mass of 0.20 \(\mathrm{kg}\) on the end of a spring oscillates with a period of 0.45 s and an amplitude of 0.15 \(\mathrm{m}\) . Find (a) the velocity when it passes the equilibrium point, (b) the total energy of the system, and (c) the equation describing the motion of the mass, assuming that \(x\) was a maximum at time \(t=0\) .

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