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Context: The use of vibration in engineering can be an effective tool to produc...

Jun 6, 2024

Context: The use of vibration in engineering can be an effective tool to produce a result. For example, some companies will use vibration between 0.1 and 100 Hz frequency to provide customers with a sense of motion either in the context of movies or in attraction park. However, vibration in the context of driving a vehicle is a problem that can lead to physical problems to the driver. One has to think of Formula One drivers having lower back pain due to uncontrolled vibration of the bottom of the car, or to truck drivers sitting in their vehicle for up to 10 hours straight. This is a well-documented domain that is still undergoing some development. Also, unwanted vibration applied to mechanical or electronic device can lead to premature failure. Therefore, it is crucial that an engineer has some knowledge of basic vibration. Problem: You are making a seat for a truck driver. The seats is on the top of a soft spring (assume the seats to be without a mass and consider only one spring, k = 1.5 kN/m), encapsulated inside a plastic wrap for security, the wrap giving a damping numerically equal to the velocity. Consider the system at rest when nobody is sitting on the seat. Assume the truck driver has a mass of 90 kg. a) What should be the minimum length of the spring for the system to actually exist? Explain. b) Write the ordinary differential equation describing the system of driver, spring and damping. c) Solve the system to find the equation of motion. d) What is the natural frequency of that system? Clearly state all your hypothesis! please answer question a) b) c) d)

Generated Graph

Solution by Steps

step 1

To determine the minimum length of the spring for the system to exist, we need to consider the equilibrium position where the spring force balances the weight of the driver

step 2

The spring force is given by Hooke's Law:

F_s = k \cdot x

, where

k = 1.5 \times 10^3 \, \text{N/m}

and

x

is the compression length

step 3

The weight of the driver is

F_g = m \cdot g

, where

m = 90 \, \text{kg}

and

g = 9.8 \, \text{m/s}^2

step 4

Setting the spring force equal to the weight:

1.5 \times 10^3 \cdot x = 90 \cdot 9.8

step 5

Solving for

x

x = \frac{90 \cdot 9.8}{1.5 \times 10^3} = 0.588 \, \text{m}

Answer

The minimum length of the spring should be 0.588 meters.

Part (b)

step 1

The ordinary differential equation (ODE) describing the system includes the mass, damping, and spring constants

step 2

The general form of the ODE is

m \cdot y''(t) + c \cdot y'(t) + k \cdot y(t) = 0

step 3

Substituting the given values:

m = 90 \, \text{kg}

c = 90 \, \text{Ns/m}

, and

k = 1.5 \times 10^3 \, \text{N/m}

step 4

The ODE becomes

90 \cdot y''(t) + 90 \cdot y'(t) + 1.5 \times 10^3 \cdot y(t) = 0

Answer

The ODE is

90 \cdot y''(t) + 90 \cdot y'(t) + 1.5 \times 10^3 \cdot y(t) = 0

Part (c)

step 1

To solve the ODE, we use the characteristic equation:

90r^2 + 90r + 1.5 \times 10^3 = 0

step 2

Dividing through by 90:

r^2 + r + \frac{1.5 \times 10^3}{90} = 0

step 3

Simplifying:

r^2 + r + 16.6667 = 0

step 4

Solving the quadratic equation:

r = \frac{-1 \pm \sqrt{1 - 4 \cdot 16.6667}}{2}

step 5

This simplifies to:

r = \frac{-1 \pm \sqrt{-65.6668}}{2} = \frac{-1 \pm i \sqrt{65.6668}}{2}

step 6

The general solution is:

y(t) = e^{-t/2} \left( C_1 \cos\left(\frac{\sqrt{65.6668}}{2} t\right) + C_2 \sin\left(\frac{\sqrt{65.6668}}{2} t\right) \right)

Answer

The equation of motion is

y(t) = e^{-t/2} \left( C_1 \cos\left(\frac{\sqrt{65.6668}}{2} t\right) + C_2 \sin\left(\frac{\sqrt{65.6668}}{2} t\right) \right)

Part (d)

step 1

The natural frequency

\omega_n

is given by

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

step 2

Substituting the given values:

k = 1.5 \times 10^3 \, \text{N/m}

and

m = 90 \, \text{kg}

step 3

\omega_n = \sqrt{\frac{1.5 \times 10^3}{90}} = \sqrt{16.6667} \approx 4.08 \, \text{rad/s}

Answer

The natural frequency of the system is approximately 4.08 rad/s.

Key Concept

Vibration Analysis in Mechanical Systems

Explanation

Understanding the dynamics of a mass-spring-damper system is crucial for designing systems that can mitigate unwanted vibrations, ensuring both comfort and longevity of mechanical components.