Question

The suspension of a car can be considered to form a mass-spring system with a undamped frequency of vibration of 0.24 Hz and a damping ratio, ζ=0.5ζ=0.5. The car is driven along a road with a series of ruts that provide a periodic, sinusoidal driving force to the suspension. If the ruts are evenly spaced with a distance 2.1 m between their crests what speed must the driver not drive at to avoid damaging the car?

Answer 1

Answer on Question #70362, Physics / Other

The suspension of a car can be considered to form a mass-spring system with a undamped frequency of vibration of $0.24\mathrm{Hz}$ and a damping ratio, $\zeta = 0.5$. The car is driven along a road with a series of ruts that provide a periodic, sinusoidal driving force to the suspension. If the ruts are evenly spaced with a distance $2.1\mathrm{m}$ between their crests what speed must the driver not drive at to avoid damaging the car?

Solution:

From given the undamped frequency

$$\omega_ {0} = 2 \pi f = 2 \pi \times 0. 2 4$$

Period of forcing is

$$T = \frac {d}{v} = \frac {2 . 1 m}{v}$$

where $\mathbf{v}$ is speed.

Hence

$$\omega = \frac {2 \pi}{T} = \frac {2 \pi v}{d}$$

The peak amplitude occurs at a frequency ratio of

$$\frac {\omega}{\omega_ {0}} = \sqrt {1 - 2 \zeta^ {2}}$$

At resonance

$$\frac {2 \pi v}{d} = 2 \pi f \sqrt {1 - 2 \zeta^ {2}}$$ $$v = f d \sqrt {1 - 2 \zeta^ {2}} = 0. 2 4 \times 2. 1 \times \sqrt {1 - 2 \times 0 . 5 ^ {2}} = 0. 3 5 6 m / s$$

Answer: $0.356 \, m/s$

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