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

Respuesta :

Answer:

Part a)

[tex]k = 6.06 \times 10^4 N/m[/tex]

Part b)

[tex]b = 1795.4 kg/s[/tex]

Explanation:

Part a)

as the mass of the suspension system is given as

[tex]m = 2100 kg[/tex]

also we have

[tex]x = 8.5 cm[/tex]

so now for force balance we have

[tex]mg = kx[/tex]

[tex](525)(9.81) = k(0.085)[/tex]

[tex]k = 6.06 \times 10^4 N/m[/tex]

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

[tex]A = 0.37 A_o[/tex]

also we know

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

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

[tex]\frac{bt}{2m} = 1[/tex]

[tex]b = \frac{2m}{t}[/tex]

here t = time period of one oscillation

so it is

[tex]t = 2\pi\sqrt{\frac{m}{k}}[/tex]

[tex]t = 2\pi\sqrt{\frac{525}{6.06 \times 10^4}}[/tex]

[tex]t = 0.58 s[/tex]

now damping constant is

[tex]b = \frac{2(525)}{0.58}[/tex]

[tex]b = 1795.4 kg/s[/tex]

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