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The suspension system of a 2000 kg automobile 鈥渟ags鈥 \(10 \mathrm{~cm}\) when the chassis is placed on it. Also, the oscillation amplitude decreases by \(50 \%\) each cycle. Estimate the values of (a) the spring constant \(k\) and (b) the damping constant \(b\) for the spring and shock absorber system of one wheel, assuming each wheel supports \(500 \mathrm{~kg}\).

Short Answer

Expert verified
Spring constant is 49000 N/m; damping constant is 439.5 Ns/m.

Step by step solution

01

Understanding the problem

We need to determine the spring constant and damping constant for the suspension system. Given that each wheel supports 500 kg, and the displacement due to the weight is 10 cm, we'll use this to find the spring constant. Additionally, the oscillation amplitude decreases by 50% each cycle due to damping.
02

Calculating the spring constant \(k\)

The weight supported by the spring for one wheel is \(mg = 500 \text{ kg} \times 9.8 \text{ m/s}^2 = 4900 \text{ N}\). The system sags 10 cm, which we convert to meters: \(0.1 \text{ m}\). Use Hooke's Law, \(F = kx\), to find \(k\):\[ k = \frac{F}{x} = \frac{4900 \text{ N}}{0.1 \text{ m}} = 49000 \text{ N/m} \]
03

Understanding damping characteristics

The system's oscillation amplitude decreases by 50% each cycle. This means the system is experiencing exponential decay due to damping. The relationship between the amplitude decrease and the damping ratio can provide us with the damping constant.
04

Calculating the damping constant \(b\)

The decrease in amplitude each cycle is given by \(e^{-\frac{b}{2m}T} = 0.5\). Use the known values to express this and solve for the damping constant \(b\). First, find the natural frequency \(\omega_0\):\[ \omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{49000}{500}} = \sqrt{98} \text{ rad/s} \]Given, \(T = \frac{2\pi}{\omega_0}\), substitute back:\[ 0.5 = e^{-\frac{b}{2 \cdot 500} \frac{2\pi}{\sqrt{98}}} \]Solving for \(b\), we have:\[ -\frac{b}{1000} \cdot \frac{2\pi}{\sqrt{98}} = \ln(0.5) \]\[ b = -1000 \cdot \frac{\sqrt{98}}{2\pi} \cdot \ln(0.5) \approx 439.5 \text{ Ns/m} \]
05

Conclusion

The spring constant for the suspension system is approximately \(49000 \text{ N/m}\) and the damping constant is approximately \(439.5 \text{ Ns/m}\).

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Key Concepts

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

Spring Constant Calculation
When analyzing a suspension system, the spring constant, denoted as \( k \), is pivotal in determining how much force is needed to compress or stretch a spring by a certain distance. To find the spring constant, we apply Hooke's Law, which relates the force exerted on a spring to its displacement: \[ F = kx \] where \( F \) is the force applied, \( x \) is the displacement, and \( k \) is the spring constant. In this problem, each wheel supports a load of \( 500 \text{ kg} \). Using gravity, we calculate the force: - \( F = 500 \text{ kg} \times 9.8 \text{ m/s}^2 = 4900 \text{ N} \) The suspension system "sags" \( 10 \text{ cm} \), which is \( 0.1 \text{ m} \). Substitute these values into Hooke's Law: \[ k = \frac{4900 \text{ N}}{0.1 \text{ m}} = 49000 \text{ N/m} \] The spring constant, therefore, is \( 49000 \text{ N/m} \), indicating the spring's firmness in the suspension system.
Damping Constant Estimation
The damping constant, \( b \), quantifies the amount of resistance that opposes the motion in a damped system. In essence, it helps in controlling oscillations and is critical for the shock absorber's performance. When there's a reduction in oscillation amplitude by 50% each cycle, we recognize an exponential decay influenced by the damping effect. The formula governing amplitude reduction is: \[ e^{-\frac{b}{2m}T} = 0.5 \] - \( m \) represents the mass (\( 500 \text{ kg} \)), and \( T \) is the period of oscillation. To determine \( b \), first calculate the natural frequency \( \omega_0 \): \[ \omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{49000}{500}} \approx \sqrt{98} \text{ rad/s} \] Next, the period \( T \) is \( \frac{2\pi}{\omega_0} \). Substitute these into the relation: \[ 0.5 = e^{-\frac{b}{1000} \cdot \frac{2\pi}{\sqrt{98}}} \] Solving for \( b \) gives: \[ b \approx -1000 \cdot \frac{\sqrt{98}}{2\pi} \cdot \ln(0.5) \approx 439.5 \text{ Ns/m} \] Thus, the damping constant is approximately \( 439.5 \text{ Ns/m} \), reflecting the system's resistance.
Hooke's Law
Hooke's Law is a fundamental principle in the physics of elastic materials. It states that the force required to extend or compress a spring by some distance \( x \) is proportional to that distance. This is described by the equation: \[ F = kx \] - \( F \) is the force applied on the spring - \( k \) is the spring constant, a measure of the spring's stiffness - \( x \) is the displacement from the spring's equilibrium positionIn the context of a vehicle suspension system, Hooke's Law provides insight into how much a spring will compress under a given load. For example, when a car chassis is placed on the suspension causing it to sag, Hooke's Law helps determine the spring constant \( k \) by measuring the force exerted (due to the weight of the car) and the resulting displacement. This forms the basis for understanding how suspension systems adjust to weights and maintain stability.
Oscillation Amplitude Damping
Oscillation amplitude damping refers to the gradual reduction in the extent of oscillations or "bounce" over time. In vehicle suspension systems, damping is crucial because it affects ride comfort and road safety by minimizing unwanted vibrations. When a system experiences damping, there is a controlled decrease in the energy of the oscillations, often indicated by a reduction in amplitude with each cycle. In this exercise, the amplitude decreases by half every cycle, highlighting the effectiveness of the shock absorbers in managing oscillations. The process is mathematically modeled using the damping constant \( b \), which helps in understanding the rate of damping. When analyzing amplitude damping, one considers several factors:
  • How quickly the amplitude diminishes
  • The materials used in the damping mechanism
  • The frequency of oscillation
This knowledge aids in enhancing the design of suspension systems to ensure they efficiently absorb shocks without causing discomfort to passengers.

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