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 kmhr-1. The car stops and the people get out. How much does the car rise on its suspension?
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Hint

What is the frequency with which the car gets jolted?
 

Solution

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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

\[ f= \frac{1}{2\pi} \sqrt {\frac{k}{m+M}} \]
implying
\[ \frac{2\pi v}{d} = \sqrt {\frac{k}{m+M}} \]
so the spring constant is
\[ k= (m+M)\times \left[ \frac{2\pi v}{d}\right] ^2 \]
When the passengers get out the downward force on the suspension drops by mg. The suspension relaxes, reducing the upward spring force to match. The amount x by which the suspension spring extends is given by kx=mg implying
\[ x = \frac{mg}{k}= \frac{m}{m+M} \times \frac{gd^2}{(2\pi v)^2} \]
This is not a friendly-looking expression. To make yourself happier with it check the units. The LHS has units of length (metres). The RHS has units
\[ \frac{kg}{kg} \times \frac{ms^{-2} m^2}{m^2s^{-2}} \rightarrow m \]
which is reassuring.

Now put in the numbers. Recognising that

\[ 16 km\, hr^{-1} = 16 \times \frac{10^3}{3600} = 4.44 ms^{-1} \]
we find
\[ x= \frac{320}{1000+320} \times 9.8 \times \frac{16}{4 \pi^2 \times 4.44^2} =5 cm \, (1sf) \]
which seems reasonable.

If you worked this through yourself, well done!