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Chapter 14: Problem 3

(1) The springs of a 1500 -kg car compress 5.0 \(\mathrm{mm}\) when its 68 \(\mathrm{kg}\) driver gets into the driver's seat. If the car goes over a bump, what will be the frequency of oscillations? Ignore damping.

Short Answer

Expert verified
The frequency of oscillation is approximately 1.46 Hz.

Step by step solution

01

Determine the Total Mass Added

First, calculate the total mass added to the car when the driver sits in it. Sum the mass of the car and the driver's mass: \[ m = 1500 \, \text{kg} + 68 \, \text{kg} = 1568 \, \text{kg} \]
02

Calculate the Spring Constant

The next step involves determining the spring constant (\( k \)) using the formula for spring compression: \[ F = k \cdot x \] where force \( F = m_\text{driver} \cdot g \). The force is provided by the driver:\[ F = 68 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 666.4 \, \text{N} \]With compression \( x = 5.0 \, \text{mm} = 0.005 \, \text{m} \), solve for \( k \):\[ k = \frac{666.4}{0.005} = 133280 \, \text{N/m} \]
03

Calculate the System's Angular Frequency

Use the formula for the angular frequency of a mass-spring system: \[ \omega = \sqrt{\frac{k}{m}} \]Plug in the values from the previous steps:\[ \omega = \sqrt{\frac{133280}{1568}} \approx 9.17 \, \text{rad/s} \]
04

Determine Frequency of Oscillation

Convert the angular frequency to the frequency of oscillation (\( f \)) using:\[ f = \frac{\omega}{2\pi} \]Thus, the frequency is:\[ f = \frac{9.17}{2\pi} \approx 1.46 \, \text{Hz} \]

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Key Concepts

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

Spring Constant
A spring constant, usually denoted as \( k \), plays a crucial role in the behavior of spring systems. It represents the stiffness of a spring, showing how much force is needed to stretch or compress it by a certain distance.

When the spring compresses, the force opposing this compression is captured by Hooke's Law, given by \( F = k \cdot x \). Here, \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement. A higher \( k \) value means a stiffer spring, requiring more force to achieve the same displacement.
  • **Unit:** The spring constant is measured in Newtons per meter (N/m).
  • **Calculation:** For example, if a driver's weight causes a spring compression and a known force is applied, you can rearrange the equation to find \( k \), as done in the exercise.
  • **Significance:** Knowing \( k \) helps in understanding the system's response to added mass and provides an insight into the dynamic behavior of a spring-mass system.
Angular Frequency
In harmonic oscillations, angular frequency \( \omega \) is a measure of how quickly an object oscillates back and forth. It provides insight into the periodic nature of oscillations.

The angular frequency is related to the spring constant \( k \) and the mass \( m \) of the system through \( \omega = \sqrt{\frac{k}{m}} \). Here, a larger angular frequency means that the system oscillates more rapidly.
  • **Unit:** Angular frequency is measured in radians per second (rad/s).
  • **Calculation:** It's determined by the square root of the spring constant divided by the mass of the system, as seen in the step-by-step solution.
  • **Role:** Angular frequency helps in further determining the frequency \( f \), which tells us the number of oscillations per second.
When the mass-spring system's characteristics change, such as by adding mass, \( \omega \) adjusts to reflect these changes, impacting how the system vibrates.
Mass-Spring System
A mass-spring system is a simple type of harmonic oscillator that consists of a mass attached to a spring. The way it behaves when subjected to an external force, or when the force changes, embodies harmonic motion.

This system's behavior is governed by the interaction of the mass and the spring constant. When disturbed from its equilibrium position, it tends to undergo oscillations, a repeating cycle of motion.
  • **Components:**
    • The Mass: The object connected to the spring, which affects the speed and amplitude of oscillations.
    • The Spring: Defines how much the spring opposes the motion; quantified by the spring constant \( k \).
  • **Oscillation:** The natural motion of the mass-spring system is a type of simple harmonic motion, characterized by a sine or cosine function.
  • **Importance:** These systems are fundamental in understanding more complex mechanical systems, like vehicles, and are used in various real-world applications such as suspension systems in cars.

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Most popular questions from this chapter

Carbon dioxide is a linear molecule. The carbon-oxygen bonds in this molecule act very much like springs. Figure \(14-43\) shows one possible way the oxygen atoms in this molecule can oscillate: the oxygen atoms oscillate symmetrically in and out, while the central carbon atom remains at rest. Hence each oxygen atom acts like a simple harmonic oscillator with a mass equal to the mass of an oxygen atom. It is observed that this oscillation occurs with a frequency of \(f=2.83 \times 10^{13} \mathrm{~Hz} .\) What is the spring constant of the \(\mathrm{C}-\mathrm{O}\) bond?

(III) A glider on an air track is connected by springs to either end of the track (Fig. \(39 ) .\) Both springs have the same spring constant, \(k,\) and the glider has mass \(M .\) (a) Determine the frequency of the oscillation, assuming no damping, if \(k=125 \mathrm{N} / \mathrm{m}\) and \(M=215 \mathrm{g} .\) (b) It is observed that after 55 oscillations, the amplitude of the oscillation has dropped to one-half of its initial value. Estimate the value of \(\gamma,\) using Eq. \(16 .(c)\) How long does it take the amplitude to decrease to one- quarter of its initial value? \(x=A e^{-\gamma t} \cos \omega^{\prime} t\)

The human leg can be compared to a physical pendulum, with a "natural" swinging period at which walking is easiest. Consider the leg as two rods joined rigidly together at the knee; the axis for the leg is the hip joint. The length of each rod is about the same, \(55 \mathrm{~cm} .\) The upper rod has a mass of \(7.0 \mathrm{~kg}\) and the lower rod has a mass of \(4.0 \mathrm{~kg} .\) ( \(a\) ) Calculate the natural swinging period of the system. ( \(b\) ) Check your answer by standing on a chair and measuring the time for one or more complete back-andforth swings. The effect of a shorter leg is a shorter swinging period, enabling a faster "natural" stride.

(II) A vertical spring with spring stiffness constant 305 \(\mathrm{N} / \mathrm{m}\) oscillates with an amplitude of 28.0 \(\mathrm{cm}\) when 0.260 \(\mathrm{kg}\) hangs from it. The mass passes through the equilibrium point \((y=0)\) with positive velocity at \(t=0 .\) (a) What equation describes this motion as a function of time? (b) At what times will the spring be longest and shortest?

(II) A fisherman's scale stretches 3.6 \(\mathrm{cm}\) when a 2.4 -kg fish hangs from it. (a) What is the spring stiffness constant and (b) what will be the amplitude and frequency of oscillation if the fish is pulled down 2.5 \(\mathrm{cm}\) more and released so that it oscillates up and down?
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