122k views
11 votes
A compact car has a mass of 1310 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that the mass is equally distributed over the four springs.

Required:
a. What is the spring constant of each spring if the empty car bounces up and down 2.0 times each second?
b. What will be the car’s oscillation frequency while carrying four 70 kg passengers?

1 Answer

10 votes

Final answer:

The spring constant for each spring of the compact car is 16,345.6 N/m. When carrying four additional 70 kg passengers, the car's oscillation frequency would be approximately 2.03 Hz.

Step-by-step explanation:

To answer the question regarding the compact car's springs and changes in oscillation frequency given certain conditions, we apply principles of physics dealing with harmonic motion and Hooke's law.

Finding the Spring Constant

The formula for the frequency (f) of a mass-spring system is given by f = (1/2π) √(k/m), where k represents the spring constant and m represents the mass. For a bouncing car with a frequency of 2.0 Hz, we can rearrange the formula to find the spring constant k. Since the mass is evenly distributed over four springs, we consider the mass on each spring to be 1310 kg / 4 = 327.5 kg.

Therefore, k = (2πf)²m = (2π × 2.0 Hz)² × 327.5 kg = 16,345.6 N/m for each spring.

Carrying Passengers

When carrying four 70 kg passengers, the total mass increases by 70 kg × 4 = 280 kg, making it 1310 kg + 280 kg = 1590 kg. The new frequency f' can be found using the same formula, but using the increased mass. The new mass per spring is 1590 kg / 4 = 397.5 kg.

The new frequency f' = (1/2π) √(k/m') = (1/2π) √(16,345.6 N/m / 397.5 kg) = 2.03 Hz

User Dorkus Prime
by
5.0k points