You are riding in an automobile of mass 3000 kg. Assuming that you are examining the oscillation characteristics of its suspension system. The suspension 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 of (a) 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 750 kg.

(a) Now there are four springs which support the entire weight of the car, assuming each member support equal weight,and the springs are identical under equilibrium condition the net upward force due spring and shock absorber system must be equal to weight of the weight of the car acting in downward direction

We know weight of a body is

W = Mg

Where M is the mass of the body and g is acceleration due to gravity

Here mass of car

M = 3000 kg

acceleration due to gravity

g = 10 ms^{-2}

so weight of the car is

W = 3000kg × 10ms^{-2}

= 3 × 10^{4}N

Now all the four springs are at different positions along four wheels i.e. in parallel so if we assume spring constant of each spring as K so net equivalent spring constant will be

K_{E} = 4K

Now when a spring is compressed it exerts restoring force in opposite direction given as

F = Kx

Where F is the restoring force, K is the spring constant and x is the compression in the spring

Here the compression is

x = 15 cm = 0.15 m

the restoring force of spring due to compression balances the weight of the car so we have

F = W

K_{E}x = Mg

4K × 0.15m = 3 × 10^{4}N

Or we get spring constant of each member as

(b) Now since it is a system in which a mass is attached to spring so on slight disturbance from mean position, the body of car will undergo simple harmonic motion due to spring, no considering each spring, the mass supported by each spring is 1/4 th of the total mass i.e.

m = M/4 = 3000Kg/4 = 750 Kg

now the time period of oscillation of each spring undergoing Simple harmonic motion is given by relation

Where m is the mass attached to spring of spring constant K

Here the spring constant of each spring is

K = 5 × 10^{4} N/m

And mass attached to each spring is

m = 750 Kg

so we have time period of oscillations as

= 0.77 s

So time period of simple harmonic oscillation or un-damped oscillation of body of car is 0.77 s

But the motion is damped and amplitude of oscillation decreases by 50% during one complete oscillation or is reduced to half of original value after each complete oscillation

So for damping, to find amplitude at any instant we have the relation

A = A_{0}e^{bt/2m}

Where A_{0} is the original amplitude of oscillation, A is the new amplitude after time t, of a particle of mass m undergoing damped oscillations with damping constant b

Here in this case If we let original amplitude be A_{0} so new amplitude A will be 50% of A_{0} or

A = A_{0}/2

Just After completion of one oscillation time is equal to time period i.e.

t = T = 0.77 s

mass attached to or supported by each spring

m = 750 Kg

so putting values we have

i.e.

taking natural Logarithm on both sides we get

(b×0.77s/1500 Kg) = ln2

b = (0.693 × 1500Kg)/0.77s

= 1350.28 Kg/s

So the damping constant b for the spring and shock absorber system of one wheel is 1350.28 Kg/s

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