Problem 79 - Physics

Question

The front spring of a car's suspension system has a spring constant of \(1.50 \times 10^{6} \mathrm{~N} / \mathrm{m}\) and supports a mass of \(215 \mathrm{~kg}\). The wheel has a radius of \(0.400 \mathrm{~m}\). The car is traveling on a bumpy road, on which the distance between the bumps is equal to the circumference of the wheel. Due to resonance, the wheel starts to vibrate strongly when the car is traveling at a certain minimum linear speed. What is this speed?

Step-by-Step Solution

Answer

The minimum speed for resonance is approximately 33.4 m/s.

Step 1: Calculate the Natural Frequency of the Spring-Mass System

First, we need to find the natural frequency of the spring-mass system. The formula for the natural frequency \( f_n \) is \( f_n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), where \( k = 1.50 \times 10^6 \, \text{N/m} \) is the spring constant and \( m = 215 \, \text{kg} \) is the mass supported by the spring. Substitute the values to find \( f_n \).

Step 2: Simplify the Natural Frequency Calculation

Insert the values into the equation: \( f_n = \frac{1}{2\pi} \sqrt{\frac{1.50 \times 10^6}{215}} \). First calculate \( \frac{1.50 \times 10^6}{215} = 6976.74 \), and then take the square root to find \( \sqrt{6976.74} \approx 83.49 \). Finally, compute \( f_n = \frac{83.49}{2\pi} \approx 13.29 \, \text{Hz} \).

Step 3: Calculate the Circumference of the Wheel

The circumference of the wheel is given by \( C = 2\pi r \), where \( r = 0.4 \, \text{m} \). Plug in the value to calculate: \( C = 2 \pi \times 0.4 = 2.51 \, \text{m} \).

Step 4: Determine Minimum Speed for Resonance

The wheel vibrates strongly due to resonance when the car's speed \( v \) causes it to hit the bumps with a frequency equal to the wheel's natural frequency. The speed is given by \( v = f_n \times C \). Substitute \( f_n = 13.29 \approx 13.3 \, \text{Hz} \) and \( C = 2.51 \, \text{m} \) to find \( v: v = 13.3 \times 2.51 = 33.4 \, \text{m/s} \).

Key Concepts

natural frequency

In mechanical systems, the natural frequency is a key concept, referring to the frequency at which a system tends to oscillate in the absence of any driving force. This frequency is crucial in understanding resonance, where a system vibrates strongly when subjected to external forces matching its natural frequency. For a simple spring-mass system, the natural frequency \( f_n \) can be calculated using the formula:

• \( f_n = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)

Here:

• \( k \) represents the spring constant
• \( m \) denotes the mass supported by the spring

By plugging in known values, like those given in the exercise (spring constant of \( 1.50 \times 10^6 \, \text{N/m} \) and mass of \( 215 \, \text{kg} \)), one can find that the natural frequency is approximately \( 13.29 \, \text{Hz} \). Recognizing this frequency helps in predicting conditions under which the system will resonate.

spring constant

The spring constant, often denoted by \( k \), is a measure of a spring's stiffness and is integral to vibration analysis. The unit of the spring constant is Newtons per meter (N/m), indicating the force required to extend or compress the spring by a unit distance. In the context of suspension systems in vehicles, the spring constant directly affects how much a spring will deform under a given load. A higher spring constant means the spring is stiffer, and thus less prone to deformation.
In our scenario, a value of \( 1.50 \times 10^6 \, \text{N/m} \) suggests a relatively stiff spring, which is typical for automotive applications where stability and support against weight are essential. Understanding the spring constant allows engineers to design systems that minimize unwanted vibrations and ensure a smoother ride.

vibration analysis

Vibration analysis is a critical process in studying how mechanical systems respond to various excitations. It involves looking at parameters like frequency, amplitude, and damping. In vehicles, improper vibration could lead to discomfort, noise, and component wear.
During vibration analysis, especially concerning resonance, engineers aim to identify and mitigate conditions where the natural frequency matches the excitation frequency, leading to large amplitude oscillations. This is particularly vital for safety and durability. In our example with the car suspension, resonance causes the wheel to vibrate intensely when the driving speed aligns with the natural frequency of the system. Proper analysis helps in adjusting design or operation conditions to reduce such risks.

linear speed

Linear speed in a rotational context refers to the rate at which an object moves along a path. For wheels, it connects rotational motion with translational movement. The linear speed \( v \) can be calculated when analyzing systems like vehicles moving over uneven roads.
In the exercise, resonance occurs when the wheel hits each bump in sync with its natural frequency. The car's linear speed can be determined using:

• \( v = f_n \times C \)

Where:

• \( f_n \) is the natural frequency
• \( C \) is the wheel's circumference

For a natural frequency of \( 13.3 \, \text{Hz} \) and a circumference of \( 2.51 \, \text{m} \), the calculated linear speed is \( 33.4 \, \text{m/s} \). This determines the specific speed at which the car triggers resonance on the bumpy road.

circumference calculation

Circumference calculation is important for understanding distances in rotational mechanics. The circumference of a wheel, calculated using the formula \( C = 2\pi r \), represents the full distance a wheel covers in one complete rotation.
For a wheel radius defined in the exercise as \( 0.4 \, \text{m} \), the circumference becomes:

• \( C = 2 \pi \times 0.4 = 2.51 \, \text{m} \)

This measurement is vital for relating rotational and translational movements. When wheels roll over bumps spaced by this distance, it can match the rotational frequency, contributing to resonance effects at certain speeds. Thus, accurately determining the wheel's circumference is fundamental for predicting and managing vibrational challenges in mechanical systems.