Problem 80 - Physics

Question

Interactive LearningWare \(10.2\) at explores the approach taken in problems such as this one. A spring is mounted vertically on the floor. The mass of the spring is negligible. A certain object is placed on the spring to compress it. When the object is pushed further down by just a bit and then released, one up/down oscillation cycle occurs in \(0.250 \mathrm{~s}\). However, when the object is pushed down by \(5.00 \times 10^{-2} \mathrm{~m}\) to point \(P\) and then released, the object flies entirely off the spring. To what height above point \(P\) does the object rise in the absence of air resistance?

Step-by-Step Solution

Answer

The object rises approximately 0.098 m above point P.

Step 1: Determine the Spring Constant

To find the spring constant, we need to first understand the period of oscillation. The period \( T \) of a mass-spring system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \), where \( m \) is the mass and \( k \) is the spring constant. Given that one oscillation cycle occurs in \( 0.250 \; \text{s} \), set \( T = 0.250 \; \text{s} \). Since \( m/k = (T/(2\pi))^2 \), we have an expression to find \( k \) once we know or assume the mass.

Step 2: Energy Conservation Principle

Apply the conservation of energy principle. The potential energy stored in the spring when compressed to point \( P \) (\( U_s = \frac{1}{2}k (0.050)^2 \)) is fully converted to kinetic energy just as it leaves the spring and subsequently to gravitational potential energy \( mgh \) at the maximum height. Set \( U_s = mgh \).

Step 3: Cancel Mass and Solve for Height

Notice that \( m \) cancels out when we compare potential energy stored in the spring and gravitational potential energy, leading to the equation \( \frac{1}{2}k (0.050)^2 = mgh \). Rearrange to solve for \( h \): \( h = \frac{k (0.050)^2}{2mg} \). With \( g = 9.8 \; \text{m/s}^2 \), substitute the expression for \( k \) into this equation.

Step 4: Calculate \( k \) Using Period

Using \( T = 0.250 \, \text{s} \) and rearranging \( T = 2\pi \sqrt{\frac{m}{k}} \), we have \( k = \frac{4\pi^2 m}{T^2} \). Substitute \( T = 0.250 \; \text{s} \): \( k = \frac{4\pi^2 m}{0.250^2} \).

Step 5: Final Calculation and Simplification

Now, substitute \( k = \frac{4\pi^2 m}{0.0625} \) into \( h = \frac{k (0.050)^2}{2mg} \) to solve for \( h \): \( h = \frac{4\pi^2 (0.050)^2}{0.0625 \times 2 \times 9.8} \). Simplify to find the height \( h \) above point \( P \).

Step 6: Compute Numerical Value

Compute \( h \) using values: \( h = \frac{4\pi^2 \times 0.0025}{0.125 \times 9.8} \approx 0.0984 \; \text{m} \). This is the height the object rises above point \( P \).

Key Concepts

Spring Constant

In a spring oscillation problem, the spring constant, denoted by \( k \), is a measure of the stiffness of the spring. It tells us how much force is needed to compress or stretch the spring by a unit length. The higher the spring constant, the stiffer the spring and the more force it requires to compress. To determine the spring constant from oscillation, we use the formula for the period of oscillation \( T \) of a mass-spring system: \[ T = 2\pi \sqrt{\frac{m}{k}} \]Here, \( m \) is the mass attached to the spring, and \( T \) is the oscillation period. By rearranging this formula, you can solve for the spring constant \( k \) as:\[ k = \frac{4\pi^2 m}{T^2} \]Knowing the oscillation period simplifies finding the spring constant when mass is known or can be estimated. In our exercise, our object completes one oscillation in 0.250 seconds. By inputting this into the formula, researchers can identify the spring constant required to compute further dynamics in the system.

Oscillation Period

The oscillation period, represented as \( T \), expresses how long it takes for one complete cycle of motion in a mass-spring system. It is directly linked to the spring constant and the mass of the object attached to the spring. The formula \( T = 2\pi \sqrt{\frac{m}{k}} \) illustrates how the period depends on the balance between mass and spring stiffness:

• As mass \( m \) increases, the period \( T \) also increases, implying slower oscillations.
• Conversely, a larger spring constant \( k \) reduces the oscillation period, resulting in faster oscillations.

In the exercise context, the given period is 0.250 seconds for one cycle. This brief duration indicates a moderate spring stiffness and mass balance where the mass is not too heavy nor the spring too soft. Understanding this relation aids in predicting system behavior when manipulating spring properties.

Energy Conservation

Energy conservation highlights the principle that the total mechanical energy in a closed system remains constant if only conservative forces are at work. In the context of our exercise, this principle implies that the energy stored in the compressed spring (potential energy) is transformed into kinetic and then gravitational potential energy as the object rises. The mathematical expression for the energy conservation in this scenario is:

• Spring potential energy at point \( P \): \( U_s = \frac{1}{2}k(0.050)^2 \)
• Gravitational potential energy at maximum height \( h \): \( mgh \)

Setting these equal simplifies the problem because it allows us to eliminate mass and solve for height using energy terms:\[ \frac{1}{2}k (0.050)^2 = mgh \]This efficiency is valuable for computing scenarios where the mass is undetermined yet consistent transformations of energy are expected.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position relative to a gravitational source, usually Earth. In spring oscillations when an object moves against gravity, GPE becomes a critical factor in energy calculations. The formula for gravitational potential energy is:\[ U_g = mgh \]where \( m \) is the mass, \( g \) is the gravitational acceleration \( (9.8 \, \text{m/s}^2) \), and \( h \) is the height above the reference point. In our exercise, it is the energy at the apex when the object leaves the spring. The spring's stored potential energy converts to kinetic energy as it decompresses, then to gravitational potential as the object rises. By equating these forms of energy, we isolate gravitational potential energy to find the maximum height \( h \) achieved above the spring's base level. This understanding helps students relate energy forms in real-world systems, reinforcing the concepts learned in physics classes.