A 1000 kg car carrying four 82 kg people travels over a "washboard” dirt road with corrugations 4.0 m apart. The car bounces with maximum amplitude when its speed is 16 km/h. when the car stops, and the people get out, by how much does the car body rise on its suspension?
5 cm
The time period between corrugations would be T=4m(16×518)910s
f=1T=109;ω=2π f=2π×109=20π9
since at this frequency the amplitude is maximum. So it got to be its natural frequency.
As we know ω=√km⇒ k=ω2m
k=(20π29)2×(1000+(4× 82))=64659.5Nm−1
when the people get out of the car the weight change is 4×82×10=3280N
we have ω=kx
3280=64659.5x
x=0.050m≈5 cm