91Ó°ÊÓ

Hooke's Law is crucial for understanding how springs work in simple harmonic motion. It states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. The formula is expressed as \( F_s = -kx \), where \( F_s \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement from the spring's relaxed length.

In our exercise, the spring is stretched by \( 0.30 \, \text{m} \)

• The spring constant \( k \) is \( 100 \, \text{N/m} \).
• The force exerted by the spring is then \( F_s = -(100 \, \text{N/m})(0.30 \, \text{m}) = -30 \, \text{N} \).

The negative sign indicates that the spring force acts in the opposite direction of the displacement, pulling the block back toward the equilibrium position. As a result, the net force on the block includes both the spring force and the gravitational force, resulting in a net upward force on the block.

The spring constant, denoted as \( k \), is a measure of a spring's stiffness. This value is essential in determining how much force is needed to stretch or compress a spring by a certain amount. In the formula for Hooke's Law, \( k \) indicates the proportionality between force and displacement.

• If \( k \) is large, the spring is stiff and requires more force to stretch.
• If \( k \) is small, the spring is more easily compressed or stretched.

In our exercise, \( k = 100 \, \text{N/m} \). This means the spring exerts \( 100 \, \text{N} \) of force for every meter of displacement. Hence, it has a moderate stiffness, resulting in a noticeable force when stretched by \( 0.30 \, \text{m} \). Understanding the spring constant helps us predict how the spring will behave under different loads and displacements.

Kinetic energy represents the energy of motion. When an object moves, it possesses kinetic energy, and in the context of a mass-spring system in simple harmonic motion, this energy reaches its maximum when the object passes through the equilibrium point. This is because the object's velocity is greatest at this point.

• The formula for the maximum kinetic energy in a spring system is \( \text{Maximum K.E.} = \frac{1}{2} k A^2 \).
• Here, \( A \) is the amplitude of motion or the maximum displacement from equilibrium, and \( k \) is the spring constant.

In the exercise, with \( k = 100 \, \text{N/m} \) and \( A = 0.30 \, \text{m} \), the maximum kinetic energy calculates to \( 4.5 \, \text{J} \). This energy transforms back and forth between kinetic and potential forms as the block oscillates, illustrating energy conservation in simple harmonic motion.

The oscillation period is the time it takes for one complete cycle of motion. In a mass-spring system, this period depends on both the mass and the spring constant. The formula for calculating the period \( T \) of oscillation is \( T = 2\pi \sqrt{\frac{m}{k}} \), where \( m \) is the mass attached to the spring and \( k \) is the spring constant.

• In the given exercise, the mass \( m \) is calculated from the weight using \( m = \frac{W}{g} = \frac{20 \text{ N}}{9.81 \text{ m/s}^2} \approx 2.04 \text{ kg} \).
• The spring constant \( k \) is \( 100 \, \text{N/m} \).

Substituting these values, the period \( T \) comes out to \( 0.897 \text{ s} \). This means the block takes approximately 0.897 seconds to complete one back-and-forth oscillation. Understanding the period is essential for predicting the timing of oscillations in harmonic motion.