Question

When a car drives along a "washboard" road, the regular bumps cause the wheels to oscillate on the springs. (What actually oscillates is each axle assembly, comprising the axle and its two wheels.) Find the speed of my car at which this oscillation resonates, given the following information: (a) When four \(80-\mathrm{kg}\) men climb into my car, the body sinks by a couple of centimeters. Use this to estimate the spring constant \(k\) of each of the four springs. (b) If an axle assembly (axle plus two wheels) has total mass \(50 \mathrm{kg}\), what is the natural frequency of the assembly oscillating on its two springs? ( \(\mathbf{c}\) ) If the bumps on a road are \(80 \mathrm{cm}\) apart, at about what speed would these oscillations go into resonance?

Short answer

Resonance occurs at approximately 18 km/h.

Step-by-step solution

Calculate Compression Due to Added Weight

4 men each having mass of 80 kg add up to 320 kg. The weight of the men is given by: \[ W = mg = 320 \times 9.8 \text{ N} = 3136 \text{ N} \]This weight compresses the springs further.

Estimate Spring Constant k

We assume the car body sinks by 2 cm under the weight of the men, which is 0.02 m. Using Hooke's Law, where force \( F = kx \), we rearrange to solve for \( k \):\[ k = \frac{F}{x} = \frac{3136}{0.02} = 156800 \text{ N/m} \]Each of the four springs supports this load, so divide by 4:\[ k_{spring} = \frac{156800}{4} = 39200 \text{ N/m} \]

Calculate the Natural Frequency of the Assembly

The axle and wheels with mass 50 kg perform simple harmonic motion. The natural frequency \( \omega_0 \) of a spring-mass system is given by:\[ \omega_0 = \sqrt{\frac{k}{m}} \]For two springs supporting the axle assembly:\[ \omega_0 = \sqrt{\frac{2 \times 39200}{50}} \approx 39.60 \text{ rad/s} \]

Corresponding Frequency and Resonance Speed

The frequency in Hz is \( f = \frac{\omega_0}{2\pi} \), thus:\[ f \approx \frac{39.60}{2\pi} \approx 6.30 \text{ Hz} \]If bumps are 0.8 m apart, the resonance speed \( v \) is:\[ v = f \times \text{wavelength} = 6.3 \times 0.8 \approx 5.04 \text{ m/s} \]

Convert Speed to km/h

To convert from meters per second to kilometers per hour, multiply by 3.6:\[ v = 5.04 \times 3.6 \approx 18.14 \text{ km/h} \]

Key concepts

Spring Constant Calculation

When you're dealing with a spring and need to determine the spring constant, you're investigating how "stiff" the spring is. The spring constant is denoted by "k" and has units of N/m, which means Newtons per meter. Let's break this down further.

You can calculate the spring constant using Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed. The equation is:

• F = kx

Here, F is the force applied to the spring, x is the displacement from its resting position, and k is the spring constant. By rearranging the formula to solve for k, you get:

• k = \( \frac{F}{x} \)

This formula tells us that if you know the force applied and the distance the spring was compressed or stretched, you can find the spring constant. In the car example, the weight of four men compressing the car's springs helps estimate the car's spring constant by using the displacement caused by their weight.

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. It's the way springs and other oscillatory systems move.

Some notable features of SHM include:

• The motion is sinusoidal in time, meaning it can be described using sine or cosine functions.
• The system oscillates around an equilibrium position.
• The total energy is conserved, being swapped between kinetic energy and potential energy.

In our scenario with the car, the oscillation of the axle assembly on the springs represents SHM. As the car moves, the wheels are subjected to periodic forces and displacements due to the "washboard" road, leading to oscillatory motion of the axle-spring system, similar to a mass bouncing on a spring.

Natural Frequency

Every oscillating system, like a spring-mass system, has a natural frequency. This is the frequency at which the system tends to oscillate in the absence of any external force. For the axle and wheels, understanding natural frequency helps predict how it oscillates.

The natural frequency, given in radians per second, is calculated for a spring-mass system using:

• \( \omega_0 = \sqrt{\frac{k}{m}} \)

In this equation, \( \omega_0 \) is the natural angular frequency, k is the spring constant, and m is the total mass of the system.

An interesting fact about natural frequency is that when an external frequency matches the natural frequency of an object, resonance occurs, which significantly amplifies the amplitude of the oscillation. This knowledge is crucial when analyzing the phenomenon of the axle assembly oscillating on a washboard road.

Hooke's Law

Hooke's Law is foundational in understanding spring behaviors and is the key to calculating the force in a linear spring. It describes the relationship between the force exerted by a spring and its extension or compression. The law is succinctly captured as:

• F = kx

Where:

• F is the force applied to the spring,
• k is the spring constant, and
• x is the displacement from the equilibrium position.

Applying Hooke's Law allows us to predict how much force a spring will exert for a given compression or extension. This makes it incredibly useful in engineering and physics applications, including our example with the car's reaction to additional weight.

Understanding Hooke's Law is essential for interpreting the behavior of the spring system under various loads and conditions, and it helps in estimating parameters like spring constant which are crucial for analyzing mechanical systems.