Understanding Natural Frequency in Passenger Cars

The natural frequency (\(\omega_n\)) of a system is the frequency at which it tends to oscillate in the absence of any driving or damping forces. For a passenger car's suspension system, understanding its natural frequency is crucial as it affects ride comfort and handling. The suspension acts like a spring-mass system.

In a simple spring-mass system, the natural frequency is determined by the stiffness of the spring and the mass attached to it.

Formulating Natural Frequency for Car Suspension

We can model the passenger car suspension as a simple spring-mass system. The suspension has a combined stiffness of \(k\) N/mm. The weight of the car is given as \(w\) Newton.

First, we need to convert the stiffness from N/mm to N/m for consistency with standard units (using metres). Since 1 mm = 0.001 m, the stiffness in N/m is:

Stiffness (\(k\)) in N/m = \(k_{\text{N/mm}} \times \frac{1 \text{ N/mm}}{0.001 \text{ m/mm}} = 1000k_{\text{N/mm}}\) N/m

However, the options provided seem to use the stiffness value directly as \(k\) in N/mm, and the standard formula is often given with \(k\) in N/m and \(m\) in kg. Let's re-examine the options and the standard formula.

The standard formula for the natural frequency (\(\omega_n\)) of a simple undamped spring-mass system in radians per second is:

\[\omega_n = \sqrt{\frac{k}{m}}\]

Where:

• \(k\) is the stiffness of the spring in N/m
• \(m\) is the mass in kg

The question gives the weight \(w\) in Newton. Weight (\(w\)) is related to mass (\(m\)) by the equation \(w = mg\), where \(g\) is the acceleration due to gravity (approximately \(9.81 \text{ m/s}^2\)). Therefore, the mass \(m\) is given by:

\[m = \frac{w}{g}\]

Now, substitute this expression for mass \(m\) into the natural frequency formula:

\[\omega_n = \sqrt{\frac{k}{m}} = \sqrt{\frac{k}{\frac{w}{g}}} = \sqrt{\frac{kg}{w}}\]

This formula uses \(k\) in N/m. If \(k\) is given in N/mm, we need to be careful. Let's assume the provided options are based on \(k\) being used with consistent units, which likely means the standard \(k\) in N/m formula is used and the options are variations.

Looking at the options provided:

1. \(\omega_n = \sqrt {\frac{1}{{km}}}\)
2. \(\omega_n = \sqrt {km}\)
3. \(\omega_n = \sqrt {\frac{k}{m}}\)
4. \(\omega_n = \sqrt {\frac{{km}}{2}}\)

Comparing these options with the standard natural frequency formula \(\omega_n = \sqrt{\frac{k}{m}}\), we can see that Option 3 directly matches the standard formula relating stiffness \(k\) and mass \(m\). This implies that the problem intends for \(k\) to be used as stiffness and \(m\) as mass, and the unit of \(k\) (N/mm) might be a detail to consider in an actual calculation, but the *form* of the formula is what's being tested.

The formula \(\omega_n = \sqrt{\frac{k}{m}}\) represents the natural frequency of a simple spring-mass system, which is a fundamental concept in vibration analysis applied to car suspension systems. Here, \(k\) would represent the equivalent stiffness of the suspension (in N/m) and \(m\) would represent the mass supported by the suspension (in kg).

The correct formula for the natural frequency (\(\omega_n\)) of a system with stiffness \(k\) and mass \(m\) is indeed \(\sqrt{\frac{k}{m}}\).

Final Formula for Natural Frequency

Based on the standard definition and the provided options, the natural frequency (\(\omega_n\)) of a passenger car's suspension system, modeled as a spring-mass system with equivalent stiffness \(k\) and mass \(m\), is given by:

\[{\omega _n} = \sqrt {\frac{k}{m}}\]

This formula is used widely in mechanical vibration analysis, including the study of vehicle suspensions.