Question

1-/7 points 1. Suppose that a car weighing 2000 pounds is supported by four shock absorbers Each shock absorber has a spring constant of 6250 lbs/foot, so the effective spring constant for the system of 4 shock absorbers is 25000 lbs/foot. 1. Assume no damping and determine the period of oscillation of the vertical motion of the car. Hint: g- 32 ft/sec2 25000.32 2. After 10 seconds the car body is 1 foot above its equilibrium position and at the high point in its cycle. What were the initial conditions? ft. and y'(0) t/sec. 3. Now assume that oil is added to each the four shock absorbers so that, together, they produce an effective damping force of 7.13 lb-sec/ft times the vertical velocity of the car body. Find the displacement y(t) from equilibrium if y0) 0 ft and y0)-10 ft/sec. 2. Suppose that you are designing a new shock absorber for an automobile. The car has a mass of 1500 kg (kilograms) and the combined effect of the springs in the suspension system is that of a spring constant of 2000 N/m (i.e.each of the four springs has a spring constant of 500 N/m) 1. Before a damping mechanism is installed in the car, when the car hits a bump it will bounce up and down. How may bounces will a rider experience in the minute right after the car hits a bump? Alternatively, what is the frequency in cycles per minute? times per minute. 2. Your job is to design a damping mechanism which eliminates oscillations when the car hits a The car will bounce bump. What is the minimum value of the effective damping constant that can be used? Y-kg/sec. 3. Suppose that at time t-0 the car hits a bump. Immediately before that time the car was not moving up and down and the effect of the bump is to add a vertical component to the speed of the car of 1.0 meter/sec. How high will the car rise above its equilibrium position if you design the system with the damping constant you found in part (b)? It will rise meters above the equilibrium position.

Answer 1

(1) (1) Calculate the period of oscillation as follows: 27T T = 27T = k = 27T k mg =27 k(g) W 2(3.14) k(g) 2000 pounds =2 (3.14) - (25000 lbs/ft) 32 ft/s =0.314 s (2) The angular velocity of car is, 2(3.14) 277 20 rad/s 0.314 s T The displacement of car is given by y (t)A cos y(10 s) (1 ft)cos ((20 rad/s) (10 s)+ - 1ft (1 ft)cos (200+¢) =1 cos(200 200 0 =-200

The initial displacement of car at t 0 s is, y(0) (1 ft)cos((0)-200) 0.48719 ft The initial velocity of car at t=0 s is, y(0)(1ft)(20 rad/s)sin ((20 rad/s) (0)-200) =-(20 ft/s)(0.8733) =-17.466 ft/s (3) The damping coefficient is bg 2mg (-7.13 lb s/ft)(32 ft/s2 2(2000 pounds) =-0.05704 rad/s The damping frequency is, (20 rad/s) -(-0.05704 rad/s) 20 rad/s The displacement of car at the time of damping is y(t) Ae cos(t + ¢) For t 0 s, we have y(0) =0.

Now y (t) Ae (0) А cos ф 3 0 cos('t+ -yt 2 Hence У () --Ае"о' sin (w't + ф)- Ае "усos (e't + ) sin (t+)+ycos (ot+ -yt -Ae[ )] y' (0)-Ae 'sin (es' (0) + ¢) +y cos(a (0)+) -10 ft/s -4(20 rad/s)sin ()+(-0.05704 rad/s) cos () -yt -10 ft/s A (20 rad/s) sin +(-0.05704 rad/s) cos 2 2 -10 ft/s -A(20 rad/s) A 0.5 ft Therefore, the damping equation is y(t)(0.5 ft)e05704 rad 3)* cos (20 rad/s)t 2