You are riding 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. g = 10 m/s2.

Certainly! As a seasoned tutor on UrbanPro, I'd approach this problem step by step, ensuring a thorough understanding for my students.

Firstly, let's address the sagging of the suspension. When the entire automobile is placed on the suspension, it sags 15 cm. This sag can be attributed to the force exerted by the weight of the automobile, which is countered by the restoring force of the spring.

The force exerted by the weight of the automobile can be calculated using the formula:

Fweight=m×gFweight=m×g

Where:

• mm = mass of the automobile = 3000 kg
• gg = acceleration due to gravity = 10 m/s²

Fweight=3000 kg×10 m/s2Fweight=3000kg×10m/s2 Fweight=30000 NFweight=30000N

Now, to find the spring constant kk, we'll use Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position:

Fspring=k×xFspring=k×x

Where:

• FspringFspring = force exerted by the spring
• kk = spring constant
• xx = displacement from equilibrium position

Given that the suspension sags 15 cm (or 0.15 m) under the weight of the automobile, we can set up the equation:

30000 N=k×0.15 m30000N=k×0.15m

From this, we can solve for kk:

k=30000 N0.15 mk=0.15m30000N k=200000 N/mk=200000N/m

So, the spring constant kk is 200000 N/m200000N/m.

Moving on to the damping constant bb, we're told that the amplitude of oscillation decreases by 50% during one complete oscillation. This indicates damping in the system.

The formula for damped oscillations involves the damping constant, and given that the amplitude decreases by 50% each oscillation, we can use this information to find bb.

The formula for the amplitude of damped oscillations is given by:

A=A0×e−b2mtA=A0×e−2mbt

Where:

• AA = amplitude at time tt
• A0A0 = initial amplitude
• bb = damping constant
• mm = mass of the system
• tt = time

Since the amplitude decreases by 50%, A=0.5A0A=0.5A0 after one complete oscillation. Substituting this into the formula:

0.5A0=A0×e−b2mT0.5A0=A0×e−2mbT

Where TT is the period of one complete oscillation.

0.5=e−b2mT0.5=e−2mbT

Taking the natural logarithm of both sides:

ln⁡(0.5)=−bT2mln(0.5)=−2mbT

We're given that the amplitude decreases by 50% during one complete oscillation. In simple harmonic motion, the period (TT) is related to the spring constant (kk) and the mass (mm) of the system:

T=2πmkT=2πkm