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Chapter 16: Problem 62

62\. Washboard Road A \(1000 \mathrm{~kg}\) car carrying four \(82 \mathrm{~kg}\) people travels over a rough "washboard" dirt road with corrugations \(4.0 \mathrm{~m}\) apart, which cause the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is \(16 \mathrm{~km} / \mathrm{h}\). The car now stops, and the four people get out. By how much does the car body rise on its suspension due to this decrease in mass?

Short Answer

Expert verified
The car body rises by 5.24 cm when the four people get out.

Step by step solution

01

鈥� Convert Speed to m/s

Convert the car's speed from km/h to m/s. Use the conversion factor: 1 km/h = 0.27778 m/s. Initial speed: \(16 \text{ km/h} = 16 \times 0.27778 \text{ m/s} = 4.4448 \text{ m/s}\)
02

鈥� Determine Bouncing Frequency

Find the frequency of the car's bounce using the distance of corrugations (4.0 m) and the car's speed (4.4448 m/s). Frequency: \(f = \frac{\text{Speed}}{\text{Distance}} = \frac{4.4448 \text{ m/s}}{4.0 \text{ m}} = 1.1112 \text{ Hz}\)
03

鈥� Calculate Effective Spring Constant

Use the relationship between mass, gravity, and spring constant for the fully loaded car. The total mass is the mass of the car plus the mass of the people. Total mass: \(m_{\text{total}} = 1000 \text{ kg} + 4 \times 82 \text{ kg} = 1328 \text{ kg}\)The formula for frequency in a mass-spring system is: \(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\). Solving for the spring constant, \(k\), we get: \[k = (2\pi f)^2 m\]Substitute the values: \[k = (2\pi \times 1.1112)^2 \times 1328\]\( = 61257.34 \text{ N/m}\)
04

鈥� Calculate New Suspension Rise

When the four people get out, the decrease in mass is: \(m_{\text{people}} = 4 \times 82 \text{ kg} = 328 \text{ kg}\)The suspension rise can be calculated using Hooke's law \(F = k\Delta x\), where \(\Delta x\) is the rise and \(F\) is the force change (weight of the people). \[\Delta x = \frac{F}{k} = \frac{m_{\text{people}} \times g}{k}\]Substitute the known values: \[\Delta x = \frac{328 \times 9.8}{61257.34} = 0.0524 \text{ m}\]\This is the rise in meters, which can also be expressed in centimeters.

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Key Concepts

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

Mass-Spring System
In physics, a mass-spring system is a basic model that shows how mass interacts with a spring. When you compress or stretch a spring, it gains potential energy.
Let it go, and this energy converts to kinetic energy, causing the mass to move. The motion is usually periodic and can be described using simple harmonic motion principles.
For example, in a car's suspension system, the car's mass acts as the 'mass', and the springs in the suspension act as the 'spring'. Understanding this system is essential to solve any bounce-related problems.
Think of it like a child鈥檚 playground swing, where the seat (mass) moves back and forth because of the chains (springs). The principles are similar!
Spring Constant Calculation
The spring constant, denoted by k, measures the stiffness of a spring. It鈥檚 a central figure in Hooke鈥檚 Law, which states that the force needed to extend or compress a spring scales linearly with the distance of that extension or compression.
The formula to find the spring constant k in a mass-spring system is: k = (2蟺f)^2 * m
Here, f is the frequency, and m is the mass being supported by the spring.
For our washboard road problem, the total mass of the car and people is 1328 kg, and the bounce frequency is 1.1112 Hz. Substituting these values, we get: k = (2蟺 * 1.1112)^2 * 1328 鈮� 61257.34 N/m
This calculation tells us how stiff the car鈥檚 suspension springs are. The higher the value, the stiffer the spring.
Frequency Determination
Frequency refers to how often an event repeats over time, such as the car bouncing up and down on the road. It is measured in Hertz (Hz), which means cycles per second.
We determine the frequency using the car speed and the distance between bumps on the road. The formula is: f = Speed / Distance
For our example, the car travels at 4.4448 m/s over bumps 4.0 meters apart, resulting in: f = 4.4448 m/s / 4.0 m = 1.1112 Hz
This means the car鈥檚 suspension system bounces back to its starting point just over once per second, creating the washboard effect.
Unit Conversion
Before solving physics problems, converting units to a standard system (usually the metric system) is vital. Here鈥檚 why:
  • Consistency: Ensures calculations are accurate.
  • Simplicity: Easy to compare different quantities.
For example, converting the car's speed from km/h to m/s helps find the bouncing frequency accurately.
The conversion factor is: 1 km/h = 0.27778 m/s
Therefore, the car鈥檚 speed: 16 km/h = 16 * 0.27778 m/s = 4.4448 m/s
Always remember, unit conversion is the first step in approaching any physics problem involving different measurement systems.
Hooke's Law
Hooke's Law describes the behavior of springs and other elastic materials. It is given by the formula: F = k螖x
Here, F is the force applied to the spring, k is the spring constant, and 螖x is the displacement from the equilibrium position.
In the washboard road problem, when the car stops and the people get out, the car鈥檚 mass decreases, and the spring鈥檚 upward force causes the car body to rise. Using Hooke鈥檚 Law, we can find this rise. Knowing the mass of the people, 螖x = (m * g) / k
Substituting the values: 螖x = (328 kg * 9.8 m/s^2) / 61257.34 N/m 鈮� 0.0524 m
This shows how understanding Hooke鈥檚 Law helps solve practical problems in everyday contexts like car suspensions.

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Most popular questions from this chapter

26\. Tuning Fork The end of one of the prongs of a tuning fork that executes simple harmonic motion of frequency \(1000 \mathrm{~Hz}\) has an amplitude of \(0.40 \mathrm{~mm}\). Find (a) the magnitude of the maximum acceleration and (b) the maximum speed of the end of the prong. Find (c) the magnitude of the acceleration and (d) the speed of the end of the prong when the end has a displacement of \(0.20 \mathrm{~mm}\).

31\. Balance Wheel The balance wheel of a watch oscillates with a rotational amplitude of \(\pi\) rad and a period of \(0.500 \mathrm{~s}\) Find (a) the maximum rotational speed of the wheel, (b) the rotational speed of the wheel when its displacement is \(\pi / 2 \mathrm{rad}\), and (c) the magnitude of the rotational acceleration of the wheel when its displacement is \(\pi / 4 \mathrm{rad}\).

12\. A Body Oscillates A body oscillates with simple harmonic motion according to the equation $$ x=(6.0 \mathrm{~m}) \cos [(3 \pi \mathrm{rad} / \mathrm{s}) t+\pi / 3 \mathrm{rad}] $$ At \(t=2.0 \mathrm{~s}\), what are (a) the displacement, (b) the velocity, (c) the acceleration, and (d) the phase of the motion? Also, what are (e) the frequency and (f) the period of the motion?

29\. Unstretched Length A uniform spring with unstretched length \(L\) and spring constant \(k\) is cut into two pieces of unstretched lengths \(L_{1}\) and \(L_{2}\), with \(L_{1}=n L_{2}\). What are the corresponding spring constants (a) \(k_{1}\) and (b) \(k_{2}\) in terms of \(n\) and \(k ?\) If a block is attached to the original spring, as in Fig. \(16-10\), it oscillates with frequency \(f\). If the spring is replaced with the piece \(L_{1}\) or \(L_{2}\), the corresponding frequency is \(f_{1}\) or \(f_{2}\). Find (c) \(f_{1}\) and \((\mathrm{d}) f_{2}\) in terms of \(f\).

13\. Piston in Cylinder The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of \(0.76 \mathrm{~m}\). If the piston moves with simple harmonic motion with a frequency of 180 rev/min, what is its maximum speed?
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