An automobile with a mass of 1000kg, including passengers, settles 1.0cm closer

Question

An automobile with a mass of 1000kg, including passengers, settles 1.0cm closer to the ground for every 100kg of additional passengers. It is driven at a constant horizontal speed of 20km/hr over a washboard road with sinusoidal bumps. The amplitude and wavelength of the bumps are 5.0cm and 20cm respectively. The distance between the front and back wheels is 2.4m.

Find the amplitude of oscillation of the car assuming it operates as an undamped driven harmonic oscillator. Neglect the mass of the wheels and assume that they are always in contact with the road.

Explanation / Answer

kx=mg

k(x-0.01)=(m+100)g

solving above two equations we get K=9.8*10^4 Nm

Frequency of bumps:

y(t)=cos(?t)

where:

?=2?f

The car travels at 5.5m/s

at 1bump/0.2m multiply it by the velocity and the frequency is 27.7 bumps/sec; hence

?=27.7?2??=1.74?10^2rad/s

and the forcing function is

y(t)=0.05kcos(174t)

y(t)=4900cos(174t)

forces acting are :

gravity

spring force

and the forcing frequency

my"=?mg+ky+4900cos(174t)

y"?W0^2y=?g+4900cos(174t)/m

where W0=sqrt(k/m)

calculating steady state solution:

y(t)=A+Bsin(174t)+Ccos(174t)

?174^2Bsin(174t)?174^2Ccos(174t)+W0^2A?W0^2Bsin(174t)?W0^2Ccos(174t)=4.9cos(174t)?g

comparing linear and trignometric terms from both side of the equation we get :

A=-W0^2/g

B=0

174^2C?W0^2C=4.9

Hence

C=4.9/(-174^2+98)

which is equal to 0.161mm

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