An automobile can be considered to be mounted on fouridentical springs as far as vertical oscillations are concerned. Thesprings of a certain car are adjusted so that the oscillations have afrequency of 3.00 Hz. (a) What is the spring constant of each springif the mass of the car is 1450 kg and the mass is evenly distributedover the springs? (b) What will be the oscillation frequency if fivepassengers, averaging 73.0 kg each, ride in the car with an even distributionof mass?

To tackle this problem, we need to apply some fundamental concepts from physics, particularly those related to harmonic motion and spring systems. Let's break it down step by step.

Understanding the System

An automobile can indeed be modeled as a system of springs. In this case, we have four identical springs supporting the car's weight. The frequency of oscillation for a mass-spring system is determined by the mass attached to the springs and the spring constant.

Part (a): Finding the Spring Constant

The formula for the frequency of oscillation (f) of a mass-spring system is given by:

f = (1/2π) * √(k/m)

Where:

• f = frequency in hertz (Hz)
• k = spring constant in newtons per meter (N/m)
• m = mass in kilograms (kg)

In this case, the total mass supported by the springs is the mass of the car, which is 1450 kg. Since there are four springs, the effective mass per spring is:

m = 1450 kg / 4 = 362.5 kg

Now, we can rearrange the frequency formula to solve for the spring constant (k):

k = (2πf)² * m

Substituting the values:

• f = 3.00 Hz
• m = 362.5 kg

Now, calculating:

k = (2π * 3.00)² * 362.5

k ≈ (18.85)² * 362.5 ≈ 355.22 * 362.5 ≈ 128,000 N/m

Thus, the spring constant of each spring is approximately 128,000 N/m.

Part (b): New Frequency with Additional Passengers

Next, we need to determine the new oscillation frequency when five passengers, each averaging 73.0 kg, are added to the car. The total mass of the passengers is:

Mass of passengers = 5 * 73.0 kg = 365 kg

Now, we add this to the original mass of the car:

Total mass = 1450 kg + 365 kg = 1815 kg

Again, since the mass is evenly distributed over the four springs, the effective mass per spring is:

m = 1815 kg / 4 = 453.75 kg

Now we can use the frequency formula again to find the new frequency:

f' = (1/2π) * √(k/m')

Substituting the known values:

• k = 128,000 N/m
• m' = 453.75 kg

Calculating the new frequency:

f' = (1/2π) * √(128,000 / 453.75)

Calculating the square root and then the frequency:

f' ≈ (1/6.283) * √(282.43) ≈ (1/6.283) * 16.81 ≈ 2.68 Hz

Therefore, the new oscillation frequency with the additional passengers is approximately 2.68 Hz.

Summary

In summary, we found that the spring constant of each spring is about 128,000 N/m, and with the added weight of the passengers, the new oscillation frequency drops to approximately 2.68 Hz. This demonstrates how the mass of a system influences its oscillatory behavior.