The front spring of a car‘s suspension system has a spring constant of 5.84 x 106 N/m and supports a mass of 273 kg. The wheel has a radius of 0.387 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?Thank you!
Well the spring-mass system exhibits simple harmonic motion with d?x/dt? -ω^2 x where ω √(k/m) 146 rad/s That corresponds to a frequency, f, of 23.3Hz and a period of T 1/f 0.0430s There will be resonance when the spring is receiving bumps at this frequency. According to the question the bumps are 2πr 2.43 meters apart. In one second the car meets 23.3 bumps and so travels 23.3*2.43 56.6 meters So the car's speed is quite fast: 56.6 m/s.
