Question

The suspension system of a 2000 kg automobile "sags" 9.9 cm when the chassis is placed on it. Also, the oscillation amplitude decreases by 39% 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 500 kg.

Im getting A correct but when solving for B, i keep getting it incorrect....
A) 49495 N/m
B) i am using A = e-bt/2m and t=2p/? and ?=vk/m

Answer 1

To find the spring constant k, use the equation k = (mg)/(∆x), where m is the mass supported by each wheel, g is the acceleration due to gravity, and ∆x is the sag distance. To find the damping constant b, use the equation b = (ln(1/e)) / (2√m√k), where e is the amplitude reduction per cycle. Substituting the given values, we find k ≈ 49494.94 N/m and b ≈ 0.1579 Ns/m.

Step-by-step explanation:

To find the spring constant k, we can use the equation:

k = (mg)/(∆x)

where m is the mass supported by each wheel (500 kg), g is the acceleration due to gravity (9.8 m/s²), and ∆x is the sag distance (0.099 m).

Substituting the values, we get:

k = (500 kg × 9.8 m/s²) / (0.099 m)

Simplifying the equation, we find that k = 49494.94 N/m.

To find the damping constant b, we can use the equation:

b = (ln(1/e)) / (2√m√k)

where e is the amplitude reduction per cycle (0.39) and m is the mass supported by each wheel (500 kg).

Substituting the values, we get:

b = (ln(1/0.39)) / (2√(500 kg × 49494.94 N/m))

Simplifying the equation, we find that b ≈ 0.1579 Ns/m.