Question

You are riding in automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspensions sags 15 cm when the entire automobile is placed on it. Also, the amplitude of oscillation decreases by 50% during one complete oscillation. Estimate the values ofa. the spring constant k and b. the damping constant b for the spring and shock absorber system of one wheel, assuming that each wheel supports.

Answer 1

(a) given, mass of the automobile, M = 3000 kg

Displacement in the suspension system, x = 15 cm = 0.15 m

There are 4 springs in parallel to the support of the mass of the automobile.

So, equation for the restoring force for the system is given by, F = –4kx = mg
Where, k is the spring constant of the suspension system
so, equivalent spring constant = 4k
Time period, T = 2π √{m/4k}

and k = mg/4x
 = (3000 × 10)/ (4 × 0.15)
= 5000 = 5 × 10⁴ Nm
Hence, spring Constant, k = 5 × 10⁴  Nm

(b) Each wheel supports a mass, m = 3000/4 = 750 kg

For damping factor b, the equation for displacement is written by $x=x_0e^{-\frac{bt}{2m}}$

The amplitude of oscilliation decreases by 50 %.
∴  x = 50 % of $x_0$ = $x_0$/2

$x_0/2=x_0e^{-\frac{bt}{2m}}$

ln2 = bt/2m

∴ b = 2m ln2/t where, t is time period.

e.g., t = 2π√{m/4k} = 2 × 3.14 √{3000/(4 × 5 × 10⁴)} = 0.7691s

Answer 2

Answer:

(a) given, mass of the automobile, M = 3000 kg

Displacement in the suspension system, x = 15 cm = 0.15 m

There are 4 springs in parallel to the support of the mass of the automobile.

So, equation for the restoring force for the system is given by, F = –4kx = mg

Where, k is the spring constant of the suspension system

so, equivalent spring constant = 4k

Time period, T = 2π √{m/4k}

and k = mg/4x

= (3000 × 10)/ (4 × 0.15)

= 5000 = 5 × 10⁴ Nm

Hence, spring Constant, k = 5 × 10⁴ Nm

(b) Each wheel supports a mass, m = 3000/4 = 750 kg

For damping factor b, the equation for displacement is written by x=x_0e^{-\frac{bt}{2m}}x=x

0

e

−

2m

bt

The amplitude of oscilliation decreases by 50 %.

∴ x = 50 % of x_0x

0

= x_0x

0

/2

x_0/2=x_0e^{-\frac{bt}{2m}}x

0

/2=x

0

e

−

2m

bt

ln2 = bt/2m

∴ b = 2m ln2/t where, t is time period.

e.g., t = 2π√{m/4k} = 2 × 3.14 √{3000/(4 × 5 × 10⁴)} = 0.7691s