B27TA Computer Lab

EXERCISE 2: Car SuspensionSTART: WEEK 3 (week begin" rel="nofollow">innin" rel="nofollow">ing 28th Sept) TO BE MARKED BY: End of computer lab session in" rel="nofollow">in Week 4 Part 1: Under certain" rel="nofollow">in circumstances, when sound travels from one medium to another, the fraction of the in" rel="nofollow">incident energy that is transmitted across the in" rel="nofollow">interface is given by E(r) = 4r /(r +1)2 where r is the ratio of the acoustic resonances of the two media. Sketch this function for r ³ 0 Without doin" rel="nofollow">ing any calculations, quickly sketch the basic shapes of the followin" rel="nofollow">ing functions. You may fin" rel="nofollow">ind these predictions useful for Exercise 3… b = e-a2 b = ae-a2 2 ( ) 4 b = a2- 1 e-a Part 2: When a sprin" rel="nofollow">ing’s motion is damped – such as in" rel="nofollow">in a car where the shock absorbers contain" rel="nofollow">in sprin" rel="nofollow">ings and either fluid or gas dampers – the motion of the end of the sprin" rel="nofollow">ing in" rel="nofollow">in response to a shock is given by: y(t) µ e-t t cos(wt) (2.1) Where t is the dampin" rel="nofollow">ing time and the angular frequency, w, is related to the shock absorber sprin" rel="nofollow">ing constant, k, as follows: m k w = (2.2) Before goin" rel="nofollow">ing further, make a sketch below predictin" rel="nofollow">ing what the basic functional form of equation 2.1 should look like – discuss your prediction with a demonstrator before proceedin" rel="nofollow">ing further. Now use Excel to in" rel="nofollow">investigate the function given by equation 2.1. Given that a typical value for the dampin" rel="nofollow">ing time in" rel="nofollow">in a car (of mass 1000 kg) is around t = 0.25 s, suggest an appropriate value for the sprin" rel="nofollow">ing constant of the shock absorber (start with a value of k = 50000 Nm-1 and work from there). Justify your conclusions. Would a heavier car require a stiffer sprin" rel="nofollow">ing?