Question

An automobile can be considered to be mounted on four springs as far as vertical oscillations are concerned. The springs of a certain car of mass \(1460 \mathrm{~kg}\) are adjusted so that the vibrations have a frequency of \(2.95 \mathrm{~Hz}\). (a) Find the force constant of each of the four springs (assumed identical). (b) What will be the vibration frequency if five persons, averaging \(73.2\) kg each, ride in the car?

Short answer

The force constant of each of the four springs is approximately \(78000 \, N/m\). If five persons, averaging \(73.2 \, kg\) each, ride in the car, the vibration frequency will be approximately \(2.24 \, Hz\)

Step-by-step solution

Calculating the Force Constant

Let's use the formula for the frequency of a simple harmonic oscillator: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\), where \(f\) is the frequency, \(k\) is the spring constant and \(m\) is the mass. This formula can be rearranged to find the spring constant \(k\): \(k = m(2\pi f)^2\). Given \(f = 2.95 \, Hz\) and \(m = 1460 \, kg\), we can calculate \(k\). Remember to divide the total spring constant by 4 at the end, as there are four springs sharing the total load.

New Frequency with Added Mass

To find the new frequency when five persons of mass \(73.2 \, kg\) each ride in the car, we first find the new total mass: \(m' = m + 5 \times 73.2\). Then, we again use the formula for the frequency of a simple harmonic oscillator: \(f' = \frac{1}{2\pi}\sqrt{\frac{k}{m'}}\), to find the new frequency \(f'\).

Plug in values to find final answers

Finally, we just need to substitute the values for \(k\), \(m\) and \(m'\) we computed in the previous two steps into the formulas. This should give us the spring constant for each spring, and the new vibration frequency.

Key concepts

Force Constant

The force constant is a fundamental concept when studying the dynamics of springs and their ability to store potential energy. In the context of simple harmonic motion, it represents the stiffness of a spring. Intuitively, think of it as a measure of how much force is needed to stretch or compress a spring by a unit of length; the greater the force constant, the stiffer the spring.

In equations, the force constant is denoted by the symbol 'k' and is a key part of Hooke's Law: \( F = -kx \), where 'F' stands for the force exerted by the spring and 'x' is the displacement from the spring's equilibrium position. The minus sign indicates that the spring's force tries to return the system to equilibrium. Therefore, knowing the force constant allows us to predict how a spring would react when forces are applied to it.

In our automobile example, once we calculate the total force constant for the car’s spring system, we can understand how it will handle the added mass of passengers.

Spring Constant

The spring constant and force constant are essentially two names for the same physical quantity and can be used interchangeably. However, the context might differ with 'spring constant' specifically referring to the characteristics of a spring.

It is directly proportional to the stiffness of the spring and inversely proportional to how much it extends or compresses for a given amount of force. Therefore, a spring with a higher spring constant 'k' is less stretchy and more resistant to deformation. The calculation of the spring constant is crucial for designing systems that must behave in predictable ways when subject to forces, an essential requirement in automotive suspension systems to provide stability and comfort.

In solving the given textbook exercise, we use the spring constant to determine how the frequency of oscillation changes with the addition of extra mass in the form of passengers, reflecting its significance in real-world applications.

Oscillation Frequency

Oscillation frequency is central to understanding simple harmonic motion. It is defined as the number of oscillations or vibrations that a system completes in one second and is measured in Hertz (Hz). The frequency of an object in simple harmonic motion like the suspension system of a car is determined by its mass and the force constant of the spring(s) it is attached to.

The relationship between these quantities is beautifully encapsulated by the formula \( f = \frac{1}{2\text{π}}\sqrt{\frac{k}{m}} \), which indicates that the oscillation frequency is inversely proportional to the square root of the mass and directly proportional to the square root of the spring constant. This elegantly shows how adding mass to the system (say, passengers in a car) will reduce the frequency of the oscillations, resulting in a less stiff ride. On the contrary, increasing the spring constant would lead to a stiffer suspension and higher oscillation frequency—useful for precision in performance vehicles.

Mass-Spring System

In physics, a mass-spring system is a classic example of simple harmonic motion where an object with a certain mass is attached to a spring, free to oscillate back and forth around an equilibrium position.

For a basic mass-spring setup, the system’s motion is determined by the mass of the object, the spring constant, and the initial displacement or initial velocity imparted to the system. This predictable behavior is what allows engineers to design car suspensions, which can be modeled as multiple mass-spring systems, ensuring that the car oscillates within desired tolerances for comfort and safety.

As the textbook exercise suggests, by calculating the frequency of vertical oscillations for a car situated on springs, we can gain insights into the suspension performance and how it would change with additional load. The principles governing a mass-spring system are not only essential for addressing academic problems but also for comprehending complex mechanical systems in practical scenarios.