Q7.19 Resonance: the downside (S)
A car of mass 103 kg carries four people of total mass 320 kg. The car travels along a road whose surface has corrugations roughly 4 m apart. The car bounces with maximum amplitude when its speed is 16 km⋅hr-1. The car stops and the people get out. How much does the car rise on its suspension?Solution
Reveal

Solution
Every time the car passes over a bump it gets a jolt; if it moves at a steady speed and the bumps are regularly spaced along the road then the jolts will be regular in time. We expect the response to these regular jolts to be largest when the frequency with which they occur is close to the natural frequency of oscillation of the car, on its suspension, satisfying the resonance condition (Key Point 6.9). If the car travels at speed v and the bumps are separated by a distance d then the time between jolts is T=d/v, and the frequency of the jolts (the number per second) is f=1/t=v/d. If this is the frequency of maximum response it is also a good estimate of the natural frequency of the car. If we model the suspension as a big spring we know that the associated frequency of oscillation of the system comprising car(mass M) and passengers (mass m) is
![\[ f= \frac{1}{2\pi} \sqrt {\frac{k}{m+M}} \]](mastermathpng-0.png)
![\[ \frac{2\pi v}{d} = \sqrt {\frac{k}{m+M}} \]](mastermathpng-1.png)
![\[ k= (m+M)\times \left[ \frac{2\pi v}{d}\right] ^2 \]](mastermathpng-2.png)
![\[ x = \frac{mg}{k}= \frac{m}{m+M} \times \frac{gd^2}{(2\pi v)^2} \]](mastermathpng-3.png)
![\[ \frac{kg}{kg} \times \frac{ms^{-2} m^2}{m^2s^{-2}} \rightarrow m \]](mastermathpng-4.png)
Now put in the numbers. Recognising that
![\[ 16 km\, hr^{-1} = 16 \times \frac{10^3}{3600} = 4.44 ms^{-1} \]](mastermathpng-5.png)
![\[ x= \frac{320}{1000+320} \times 9.8 \times \frac{16}{4 \pi^2 \times 4.44^2} =5 cm \, (1sf) \]](mastermathpng-6.png)
If you worked this through yourself, well done!