The front spring of a car‘s suspension system has a spring constant of 1.57 x 10^6 N/m and supports a mass of 205 kg. The wheel has a radius of 0.383m. The car is traveling on a bumpy road, on which the distance between 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?
Catarthur has missed pi out of the equation. Natural frequency of spring oscillations 1/ 2pi * sqrt(k/m) 13.93 Hz Circumference of wheel 2pi*0.383 2.41m Linear speed angular frequency * circumference 13.93 * 2.41 33.57 m/s
The resonant angular frequency of a mass on a spring is (omega) sqrt (k/m) The resonant frequency is (omega) / 2(pi) Find the distance between bumps circumference of wheel 2 (pi)(radius). Speed at which you'll get resonance (distance)(frequency) They give you k, m, and the radius, so you can plug and chug your way through this. Good luck!
f 1/(2Pi)sqrt(k/m) f 1/(2Pi) sqrt(1 570 000/205) 13.9 Hz The wheel has circumference C 2PiR 2.41 m At 13.9 revolution per second (hz) this represents a linear speed of v 13.9 * 2.41 33.5 m/s 120 km/h
