The front spring of a car‘s suspension system has a spring constant of 5.33 x 106 N/m and supports a mass of 296 kg. The wheel has a radius of 0.402 m. The car is traveling on a bumpy road, on which the distance between the bumps is equal to the circumference of the wheel. Due to resonance, the wheel starts to vibrate strongly when the car is traveling at a certain minimum linear speed. What is this speed?
The resonant frequency is W sqrt(k/m) or f sqrt(k/m) / (2*pi) So the suspension will be at resonance when the bumps come at a frequency of f sqrt(5.33 x 106 N/m/296 kg) / (2*pi) The period is given by T 1/f So the suspension will be at resonance when the bumps come every T seconds. The circumference of the tires is given by Circ 2*pi*r The bumps are that far apart, so the car needs to cover that distance every T seconds to set the car in resonance. V Circ/T
