Problem 70 - Physics

Question

Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{~N} / \mathrm{m}\). Four identical passengers sit down inside the car, and it is set into a vertical oscillation that has a period of \(0.370 \mathrm{~s}\). If the mass of the empty car is \(1560 \mathrm{~kg}\), determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.

Step-by-Step Solution

Answer

Each passenger has a mass of approximately 61.05 kg.

Step 1: Define the Total System Mass

The total system mass includes the mass of the car and the mass of all four passengers. Let's denote the mass of each passenger as \( m_p \). Therefore, the total mass is \( m_{total} = m_{car} + 4m_p \), where \( m_{car} = 1560 \text{ kg} \).

Step 2: Compute Total Spring Constant

Since the car is supported by four identical springs in parallel, the equivalent spring constant for the entire system is the sum of all four spring constants. Thus, \( k_{total} = 4 \times k = 4 \times 1.30 \times 10^{5} \text{ N/m} = 5.20 \times 10^{5} \text{ N/m} \).

Step 3: Apply the Formula for Oscillation Period

The formula relating the period of oscillation \( T \), mass \( m \), and spring constant \( k \) is given by the formula for a mass-spring system: \( T = 2\pi \sqrt{\frac{m}{k}} \).

Step 4: Solve for Total Mass

Rearrange the formula to solve for \( m \): \( m = \frac{kT^2}{4\pi^2} \). Substitute the known values: \( k = 5.20 \times 10^{5} \text{ N/m} \) and \( T = 0.370 \text{ s} \). Compute the total mass \( m_{total} \).

Step 5: Calculate Mass of Each Passenger

Once \( m_{total} \) is found, use the equation from Step 1: \( m_{total} = m_{car} + 4m_p \). Substitute \( m_{car} = 1560 \text{ kg} \) and solve for \( m_p \): \( m_p = \frac{m_{total} - m_{car}}{4} \).

Step 6: Perform the Calculations

Calculate \( m_{total} = \frac{5.20 \times 10^{5} \times (0.370)^2}{4\pi^2} = \frac{5.20 \times 10^{5} \times 0.1369}{39.4784} \approx 1804.2 \text{ kg}\). Then, calculate the mass of each passenger: \( m_p = \frac{1804.2 - 1560}{4} \approx 61.05 \text{ kg} \).

Key Concepts

Spring Constant

When discussing harmonic motion, the spring constant is a key element. It represents the stiffness of a spring and is denoted by the letter \( k \). The spring constant is measured in newtons per meter (N/m) and indicates the force required to compress or extend a spring by a unit length.
In the case of the automobile discussed, the car is positioned on four identical springs. Each spring has a spring constant of \( 1.30 \times 10^5 \, \text{N/m} \). Since the car uses all four springs to support its weight, these springs act in parallel.

• When springs are in parallel, the total spring constant is the sum of each spring's constant.
• Calculation: \( k_{\text{total}} = 4 \times 1.30 \times 10^5 = 5.20 \times 10^5 \, \text{N/m} \).

Understanding the spring constant is crucial, as it directly influences the car's oscillation dynamics.

Oscillation Period

The oscillation period of a system refers to the time it takes for one complete cycle of motion. In the spring mass system, this is influenced by both the total mass and the spring constant. The formula that connects these elements is:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
This equation implies that the period (\( T \)) of oscillation is determined by the mass \( m \) and the spring constant \( k \).

• A higher mass or a lower spring constant will result in a longer oscillation period.
• Conversely, increasing the spring constant or reducing the mass will shorten the period.

In our example, with an oscillation period of \( 0.370 \, \text{s} \), this relationship helps us solve for the mass involved, demonstrating how fundamental parameters govern the behavior of oscillating systems.

Mass Calculation

Determining the mass of a system in motion, especially involving oscillations, begins with understanding the forces at play. With the car's system, we use the formula for the oscillation period rearranged to solve for total mass \( m \):
\[ m = \frac{kT^2}{4\pi^2} \]
Given \( k = 5.20 \times 10^5 \, \text{N/m} \) and \( T = 0.370 \, \text{s} \), we substitute these values into the formula:
\[ m = \frac{5.20 \times 10^5 \times (0.370)^2}{4\pi^2} \approx 1804.2 \, \text{kg} \]
This calculation yields the total mass of the car and passengers combined. With the car's mass known (1560 kg), we can solve for the passenger's total mass by subtracting the car's weight:

• Each passenger's mass \( m_p = \frac{m_{\text{total}} - m_{\text{car}}}{4} \),
• Resulting in \( m_p \approx 61.05 \, \text{kg} \) per passenger.

Grasping this calculation helps in comprehending how systems maintain equilibrium and responds to dynamic forces.