Problem 68 - University Physics with Modern Physics

Question

Four passengers with combined mass 250 kg compress the springs of a car with worn-out shock absorbers by 4.00 \(\mathrm{cm}\) when they get in. Model the car and passengers as a single body on a single ideal spring. If the loaded car has a period of vibration of 1.92 s, what is the period of vibration of the empty car?

Step-by-Step Solution

Answer

The period of vibration for the empty car is shorter than 1.92 s.

Step 1: Understand the Problem

We have a car and passengers modeled as a single mass on a spring. The car compresses the spring when passengers get in, and we're given the period of vibration for this loaded car, which includes the passengers' mass. We need to determine the period of vibration of the car when empty, removing the passengers' mass.

Step 2: Identify Known Variables

The mass of the passengers is 250 kg, the displacement (compression) is 4.00 cm (0.04 m), and the period of vibration of the loaded car is 1.92 s. We need to find the period of vibration for just the car.

Step 3: Calculate the Spring Constant

Using Hooke's Law, calculate the spring constant \( k \). The force exerted by the passengers is \( F = mg \), where \( m = 250 \) kg and \( g = 9.8 \) m/s². Solve for \( k \) using the equation \( F = kx \), where \( x = 0.04 \) m. This results in \( k = \frac{mg}{x} = \frac{250 \times 9.8}{0.04} \) N/m.

Step 4: Determine Total Mass of Loaded Car

The loaded car's period \( T = 1.92 \) s gives us an expression involving the total mass \( m + M \) (where \( M \) is the mass of the car) using the formula for the period \( T = 2 \pi \sqrt{\frac{m+M}{k}} \). Rearrange to solve for \( m + M \).

Step 5: Rearrange for Car's Mass

From Step 3, solve \( m + M = \frac{kT^{2}}{4\pi^{2}} \). With the previously calculated \( k \) and \( T = 1.92 \), find \( m + M \). Then subtract the known mass of passengers \( m = 250 \) kg to find the mass \( M \) of the car.

Step 6: Calculate Period of Empty Car

With the mass of the car \( M \) known, find the period of the empty car using \( T_{empty} = 2 \pi \sqrt{\frac{M}{k}} \). Substitute \( M \) and \( k \) to find the new period.

Key Concepts

Vibration Period

The vibration period of a system, such as a loaded car modeled as a mass on a spring, describes how long it takes for one complete oscillation of the system. When the car is loaded, this period is 1.92 seconds. It means that with the added mass of passengers, the time for oscillation is longer due to increased inertia. To calculate the vibration period of an empty car, follow:

• Determine the spring constant using external forces.
• Find the effective mass during the loaded period using the period formula.
• Use the effective mass of the car alone to find the period of the empty car using the same period formula.

Remember, a stiffer spring or lighter mass decreases the vibration period, making oscillations quicker.

Hooke's Law

Hooke’s Law is fundamental when dealing with springs and describes the relationship between force and spring displacement. The law is expressed as \( F = kx \), where:

• \( F \) is the force applied to the spring.
• \( k \) is the spring constant, indicating spring stiffness.
• \( x \) is the displacement caused by the force.

In our exercise, Hooke's Law is used to find the spring constant. The force exerted by the passengers is calculated by multiplying their mass by gravitational acceleration: \( F = 250 \times 9.8 \) N. This force is then used to determine the spring constant \( k \) using displacement \( x = 0.04 \) m. Calculating \( k \) provides insight into how the spring responds to the mass and implies the potential for vibration.

Car Mass Calculation

Finding the car's mass involves a few essential steps. First, use the known vibration period of the loaded car (1.92 seconds) and the spring constant, calculated using Hooke's Law, in the formula for the period of a mass-spring system:\[T = 2 \pi \sqrt{\frac{m + M}{k}}\]Here, \( m \) is the passenger mass, and \( M \) is the car's mass. By rearranging this formula, one can solve for the total mass \( m + M \). Subtract the passenger mass from this sum to isolate \( M \). Finally, with the car's mass alone, the period of the empty car is recalculated. Repeat the process using its mass and known spring constant:\[T_{empty} = 2 \pi \sqrt{\frac{M}{k}}\]This altered period will typically be less since the car is now lighter, showcasing the dynamic interplay between mass and vibration in mechanical systems.