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STatiana [176]
3 years ago
14

The suspension system of a 2100 kg automobile "sags" 8.5 cm when the chassis is placed on it. Also, the oscillation amplitude de

creases by 63% each cycle. Estimate the values of (a) the spring constant k and (b) the damping constant b for the spring and shock absorber system of one wheel, assuming each wheel supports 525 kg.
Physics
1 answer:
Gennadij [26K]3 years ago
7 0

Answer:

Part a)

k = 6.06 \times 10^4 N/m

Part b)

b = 1795.4 kg/s

Explanation:

Part a)

as the mass of the suspension system is given as

m = 2100 kg

also we have

x = 8.5 cm

so now for force balance we have

mg = kx

(525)(9.81) = k(0.085)

k = 6.06 \times 10^4 N/m

Part b)

Now we know that amplitude decreases by 63% in each cycle

so after one cycle the amplitude will become 37% of initial amplitude

so it is given as

A = 0.37 A_o

also we know

A = A_o e^{-bt/2m}

0.37 A_o = A_o e^{-bt/2m}

\frac{bt}{2m} = 1

b = \frac{2m}{t}

here t = time period of one oscillation

so it is

t = 2\pi\sqrt{\frac{m}{k}}

t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}

t = 0.58 s

now damping constant is

b = \frac{2(525)}{0.58}

b = 1795.4 kg/s

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