91Ó°ÊÓ

In a spring-mass system, the oscillation frequency is how quickly the mass moves back and forth on the spring. Generally, it's measured in hertz (Hz), which represents the number of oscillations per second. For our given problem, the mass initially oscillates with a frequency of 0.83 Hz.

The formula connecting frequency ( \( f \) ) with spring constant ( \( k \) ) and mass ( \( m \) ) is:

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

This elegantly shows that frequency is inversely proportional to the square root of the mass. Basically, as the mass increases, the frequency decreases along a specific curve.

When an additional 780 g (or 0.78 kg) is added, the frequency reduces to 0.60 Hz. This behavior aligns with the inverse relationship between mass and frequency. In practice, this means heavier objects tend to oscillate slower since more force is needed to initiate and maintain their motion.

The spring constant ( \( k \) ) is a critical factor in determining how "stiff" a spring is. It is measured in newtons per meter (N/m) and provides insight into the relationship between the force applied to the spring and how much it stretches or compresses.

Using the frequency formula, \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), we can see \( k \) also influences the frequency. To solve problems like the one at hand, expressing \( k \) in terms of \( m \) through equation manipulation is vital. For example, as derived from the given frequencies, \( k \) can be expressed for different states of mass:

• Initial: \( k = m (2\pi \times 0.83)^2 \)
• With additional mass: \( k = (m + 0.78) (2\pi \times 0.60)^2 \)

Understanding that \( k \) remains constant allows for comparison and solving of these two equations for \( m \). This highlights \( k \)'s robustness—it doesn't change just because the mass or frequency does.

Breaking down the task of solving for unknown variables is a key skill in physics. Solving equations requires a systematic approach, especially in dynamics problems like this one.

For our scenario, derive from the given mathematical equations, both describing the original and altered system frequencies:

• First equation (for 0.83 Hz): \( \frac{k}{m} = (2\pi \times 0.83)^2 \)
• Second equation (for 0.60 Hz): \( \frac{k}{m + 0.78} = (2\pi \times 0.60)^2 \)

By squaring both sides of each equation and solving for \( k \) , one can establish a relational equality. Since \( k \) is the same for both masses, the expressions for \( k \) in terms of \( m \) are set equal. This simplifies to a single equation: \( m (2\pi \times 0.83)^2 = (m + 0.78) (2\pi \times 0.60)^2 \) .

Finally, isolating \( m \) by doing some algebraic rearrangements gives \( m = \frac{0.2808}{0.3289} \) , solving to get \( m \approx 0.854 \) kg. Throughout, each step hinges on keeping equations balanced and isolating variables logically, reinforcing fundamental mathematical and physical principles.