# The suspension system of a 2240 kg automobile "sags" 13 cm when the chassis is placed on it....

## Question:

The suspension system of a 2240 kg automobile "sags" 13 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 40% each cycle. Assume each wheel supports 560 kg.

(a) Find the value of the spring constant k for the spring and shock absorber system of one wheel.

(b) Find the damping constant b for the spring and shock absorber system of one wheel.

## Stiffness of the Spring:

Spring is a component of the suspension system used to absorb shocks and vibrations. The magnitude of the energy that can be stored in the spring during compression is directly proportional to the stiffness of the spring.

We are given the following data:

• Mass supported by each wheel, {eq}m=560\ \rm kg {/eq}
• Deflection of the suspension system, {eq}x=13\ \rm cm {/eq}

#### Question (a)

As per Hooke's law, the following equation expresses the relationship between the deflection of spring, force applied on the spring, and the spring constant:

{eq}\begin{align} F&=kx&\left [F=mg \right ]\\[0.3 cm] mg&=kx \end{align} {/eq}

• m' is the mass supported by each wheel
• k is the spring constant

Substituting values in the above equation, we have:

{eq}\begin{align} mg&=kx\\[0.3 cm] 560\ \text{kg}\times9.80\ \text{m/s}^{2}&=k\times 13\times10^{-2}\ \rm m\\[0.3 cm] 560\ \rm\dfrac{N}{m/s^{2}}\times9.80\ \text{m/s}^{2}&=k\times 13\times10^{-2}\ \rm m\\[0.3 cm] k&=\dfrac{560\times9.80\ \rm N}{13\times10^{-2}\ \rm m}\\[0.3 cm] &=42215\ \rm N/m\\[0.3 cm] &\approx\boxed{\color{blue}{42\times10^{3}\ \rm N/m}} \end{align} {/eq}

#### Question (b)

We are given that the oscillation amplitude decreases by 40% each cycle, If {eq}(x) {/eq} defines initial amplitude and {eq}(x_{n}) {/eq} defines amplitude after one cycle, we have:

{eq}x_{n}=0.40x {/eq}

The relationship between the damping ratio, damping constant, and the period of oscillation is expressed by the following equation:

{eq}\dfrac{x_{n}}{x}=\left (e \right )^{\dfrac{-bT}{2m}} {/eq}, where

• T is the period of oscillation

The period of oscillation is expressed by the following equation:

{eq}\begin{align} T&=2\pi\sqrt{\dfrac{m}{k}}\\[0.3 cm] &=2\pi\sqrt{\dfrac{560\ \rm N/(m/s)^{2}}{ 42215\ \rm N/m}}\\[0.3 cm] &=0.723\ \rm s \end{align} {/eq}

Substituting known and calculated values in the above equation, we have:

{eq}\begin{align} \dfrac{x_{n}}{x}&=e^{-bT/2m}\\[0.3 cm] 0.4&=e^{\left ( \dfrac{-bT}{2m}\right )}\\[0.3 cm] \ln 0.4&=\ln\left (e^{\left ( \dfrac{-bT}{2m}\right )} \right )\\[0.3 cm] -0.916&=\dfrac{-bT}{2m}\\[0.3 cm] b&=\dfrac{0.916\times2\times 560\ \rm kg}{0.723\ \rm s}\\[0.3 cm] &=1419.4\ \rm kg/s\\[0.3 cm] &\approx\boxed{\color{blue}{1420\ \rm kg/s}} \end{align} {/eq}

Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
3.4K

In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.