91Ó°ÊÓ

Hooke's Law is a fundamental principle that describes the behavior of springs. It states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The mathematical expression of Hooke's Law is given by the formula \[ F = kx \]where:

• \( F \) is the force exerted by the spring, measured in newtons (N)
• \( k \) is the spring constant, a measure of the stiffness of the spring, measured in newtons per meter (N/m)
• \( x \) is the displacement from the spring's equilibrium position, measured in meters (m)

In our original exercise, a 4.00 kg block caused the spring to stretch by 16.0 cm, or 0.16 meters when converted to SI units. The gravitational force on the block is \[ F = mg = 4.00 \, \mathrm{kg} \times 9.81 \, \mathrm{m/s}^2 \]Solving this provides the force, which when used in Hooke's law formula allows us to calculate the spring constant \( k \). By plugging the values into the formula, we find that the spring constant is approximately \[ k = 245.25 \, \mathrm{N/m} \].This constant tells us how much force is needed to stretch or compress the spring by one meter. A larger value of \( k \) indicates a stiffer spring.

The oscillation period of a spring-mass system is a crucial concept when studying harmonic motion. The period is the time it takes for one complete cycle of motion. When a spring oscillates, it moves back and forth around an equilibrium point in a regular pattern. The period of oscillation is given by the formula:\[ T = 2\pi\sqrt{\frac{m}{k}} \]where:

• \( T \) is the period, measured in seconds
• \( m \) is the mass attached to the spring, measured in kilograms (kg)
• \( k \) is the spring constant, measured in newtons per meter (N/m)

In the given exercise, the spring constant \( k \) was previously calculated to be 245.25 N/m. After replacing the 4.00 kg block with a 0.500 kg body, the new period of oscillation can be determined. By substituting \( m = 0.500 \) kg and the calculated \( k \) into the formula, we find that:\[ T = 2\pi\sqrt{\frac{0.500}{245.25}} \]This calculation results in a period of approximately 0.282 seconds. This means that the lighter mass takes a little over a quarter of a second to complete one full oscillation cycle. Thus, a lighter mass oscillates faster on the same spring than a heavier one.

A spring-mass system is a fundamental model in physics that demonstrates simple harmonic motion. It consists of a mass attached to a spring, which can be displaced and will oscillate back and forth due to the restoring force provided by the spring. In essence, it provides key insights into periodic motions observed in many physical systems.
To understand the dynamics of such systems, the spring constant \( k \) and the mass \( m \) play pivotal roles:

• The spring constant \( k \) represents the stiffness of the spring.
• The mass \( m \) influences how quickly or slowly the oscillation occurs.
• The period of oscillation \( T \) links both these properties and highlights how the system behaves in motion.

In practical scenarios, altering either the mass or the spring's properties can affect the frequency and oscillation period. For example, increasing the mass will typically result in slower oscillations, while increasing the stiffness (higher \( k \)) tends to make the system oscillate quicker. The exercise provided serves as a clear illustration of these principles by showing the changes in period when mass is altered. This framework not only helps in solving problems related to oscillations but also applies to various scientific and engineering contexts, such as designing suspension systems in vehicles.