Car Spring Oscillations: Frequency When Driving Over Bumps Explained!

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Consider the given data as follows.

The mass of the spring is M = 1700 kg.

The mass of the driver is m = 66 kg.

Compressed displacement by the spring is x = 5.0 mm.

The spring force is expressed as follows: F = -kx

Here, k represents the spring constant and x represents the compressed displacement.

The frequency of oscillation is expressed as follows:

\(f=\dfrac{1}{2\pi }\sqrt{\dfrac{k}{m}}\ ............\left( 1 \right)\)

Here, k is the spring constant and m is the system's mass.

Again, the spring constant is expressed as follows:

\(k=\dfrac{m\cdot g}{x}\)

Substituting the values of 66 kg for m, 5.0 mm for x, and \(9.8\ \dfrac{\text{m}}{{{\text{s}}^{\text{2}}}}\) for g in the above equation

 \(k=\dfrac{\left( 66\ \text{kg} \right)\cdot 9.8\ \dfrac{\text{m}}{{{\text{s}}^{\text{2}}}}}{5.0\ \text{mm}} \) 

 \( k=\dfrac{\left( 66\ \text{kg} \right)\cdot 9.8\ \dfrac{\text{m}}{{{\text{s}}^{\text{2}}}}}{\left( 5.0\ \text{mm} \right)\left( \dfrac{{{10}^{-3}}\ \text{m}}{1\ \text{mm}} \right)} \) 

 \( k=129,360\ \dfrac{\text{N}}{\text{m}} \) 

The driver sits on the seat in such a way that the springs of the seat are displaced lower. The spring force will exert upward force. The driver's gravitational pull acts in the downward direction. As a result, the gravitational force balances the spring force.