1.3 Simple harmonic motion: a special periodic motion  (Page 2/7)

Calculate the frequency and period of oscillations: bad shock absorbers in a car

If the shock absorbers in a car go bad, then the car will oscillate at the least provocation, such as when going over bumps in the road and after stopping (See [link] ). Calculate the frequency and period of these oscillations for such a car if the car’s mass (including its load) is 900 kg and the force constant ( $k$ ) of the suspension system is $6\text{.}\text{53}\times {\text{10}}^4\text{N/m}$ .

Strategy

The frequency of the car’s oscillations will be that of a simple harmonic oscillator as given in the equation $f=\frac{1}{\mathrm{2\pi }}\sqrt{\frac{k}{m}}$ . The mass and the force constant are both given.

Solution

1. Enter the known values of k and m :
   $f=\frac{1}{\mathrm{2\pi }}\sqrt{\frac{k}{m}}=\frac{1}{\mathrm{2\pi }}\sqrt{\frac{6\text{.}\text{53}\times {\text{10}}^4\text{N/m}}{\text{900}\text{kg}}}.$
2. Calculate the frequency:
   $\frac{1}{\mathrm{2\pi }}\sqrt{\text{72.}6/{\text{s}}^{\mathrm{–2}}}=1\text{.}{\text{3656}/\text{s}}^{\text{–1}}\approx 1\text{.}{\text{36}/\text{s}}^{\text{–1}}=\text{1.36 Hz}.$
3. You could use $T=\mathrm{2\pi }\sqrt{\frac{m}{k}}$ to calculate the period, but it is simpler to use the relationship $T=1/f$ and substitute the value just found for $f$ :
   $T=\frac{1}{f}=\frac{1}{1\text{.}\text{356}\text{Hz}}=0\text{.}\text{738}\text{s}.$

Discussion

The values of $T$ and $f$ both seem about right for a bouncing car. You can observe these oscillations if you push down hard on the end of a car and let go.