91Ó°ÊÓ

A mass-spring system consists of a spring attached to a mass, which is free to move. This system can oscillate back and forth, forming what is known as simple harmonic motion (SHM) when displaced from its equilibrium position. In the context of this exercise, imagine a spring hanging vertically with a mass attached at its end. When you pull the mass down and release it, the system enters into SHM, moving up and down repeatedly.
The characteristics of this motion depend on the properties of the spring and the mass. It's important to understand that the mass affects how quickly or slowly the system oscillates. The spring's elasticity, defined by its spring constant, also plays a crucial role. Together, these elements determine the system’s unique oscillatory behavior.

The frequency of oscillation in a mass-spring system tells us how many cycles of the motion occur in a second. It is denoted by the symbol \( f \) and is measured in Hertz (Hz). In our exercise, one mass causes the system to oscillate at 12.0 Hz, meaning it completes 12 oscillations every second.
Importantly, the frequency depends on both the mass and the spring constant. You can calculate it using the formula:

• \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \)

Here, \( k \) is the spring constant, and \( m \) is the mass. This equation shows how frequency decreases as mass increases; adding more mass makes the oscillations slower.

The spring constant, represented by the letter \( k \), is a measure of a spring's stiffness. Physically, it reflects how hard you need to pull or push to change the spring's length. In a mass-spring system, the spring constant combines with the mass to determine the frequency of oscillation.
In the provided exercise, the step-by-step solution involves deriving the spring constant \( k \) using the known mass \( m_1 \) and frequency \( f_1 \) of 12.0 Hz. The essential formula used is

• \( k = (2\pi f)^2 m \)

which shows that the spring constant directly affects how quickly the spring returns to equilibrium after being stretched or compressed. Understanding \( k \) allows us to predict how the system's properties will change as we add more mass.

Calculating the mass ratio is the crux of solving this exercise. The ratio \( m_2/m_1 = 8 \) specifies how much larger the second mass is compared to the first. This was derived from the frequencies of oscillation before and after adding the second mass.
We began by expressing the spring constant \( k \) for the original mass \( m_1 \) and used it to determine the effect on frequency when a new mass \( m_2 \) was added. The system's frequency dropping to 4.00 Hz indicates that additional mass slowed the oscillations, thus allowing us to set up an equation to relate the two frequencies:

• \( 16 (m_1 + m_2) = 144 m_1 \)

Simplifying this equation gives the mass ratio, providing a clear insight into how additional mass affects the oscillation of the spring.