# When a 88.0 kg person climbs into an 1500 kg car, the car's springs compress vertically 1.1 cm....

## Question:

When a {eq}88.0\ \rm{kg} {/eq} person climbs into an {eq}1500\ \rm{kg} {/eq} car, the car's springs compress vertically {eq}1.1\ \rm{cm} {/eq}. What will be the frequency of vibration when the car hits a bump? (Ignore damping.)

## Spring-mass System:

The car's suspension system has four compression springs, and they all take up the compression when the car hits a bump. The frequency of the car's vibration depends on the mass of the car and the mass of the occupants sitting in the car, and the equivalent spring constant of the springs. When a car is empty, we experience more vibration than a loaded car.

Given data:

• {eq}m=\rm 88.0 \ kg {/eq} is the mass of the person
• {eq}M=\rm 1500 \ kg {/eq} is the mass of the car
• {eq}x=\rm 1.1 \ cm=0.011 \ m {/eq} is the compression of springs
• {eq}k {/eq} is the equivalent spring constant
• {eq}f {/eq} is the frequency of oscillation

The weight of the person causes the springs to compress by {eq}\rm 0.011 \ m {/eq}. Using Hooke's law:

\begin{align} F&=kx \\[0.3cm] mg&=kx \\[0.3cm] k&=\dfrac{mg}{x} \\[0.3cm] &=\rm \dfrac{88.0 \ kg \times 9.81 \ \frac{m}{s^2}}{0.011 \ m} \\[0.3cm] &=\rm 78480 \ N/m \end{align}

Therefore, the equivalent spring constant is {eq}\rm 78480 \ N/m {/eq}.

The frequency of oscillation is given by:

\begin{align} f&=\dfrac{1}{2\pi} \sqrt{\dfrac{k}{m+M}} \\[0.3cm] &=\rm \dfrac{1}{2\pi} \sqrt{\rm \dfrac{78480 \ \frac{N}{m}}{(88.0 \ kg +1500 \ kg )}} \\[0.3cm] &\approx \color{blue}{\boxed { \rm 1.1 \ Hz}} \ \ \ \ \ \rm (correct \ to \ two \ significant \ figures ) \end{align}

Therefore, the frequency of vibration is {eq}\mathbf{ 1.1 \ Hz} {/eq}.

Practice Applying Spring Constant Formulas

from

Chapter 17 / Lesson 11
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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.