Question

A 1.00-kg mass is suspended vertically from a spring with \(k=100 . \mathrm{N} / \mathrm{m}\) and oscillates with an amplitude of \(0.200 \mathrm{~m}\). At the top of its oscillation, the mass is hit in such a way that it instantaneously moves down with a speed of \(1.00 \mathrm{~m} / \mathrm{s}\). Determine a) its total mechanical energy, b) how fast it is moving as it crosses the equilibrium point, and c) its new amplitude.

Short answer

Question: A 1.00 kg mass-spring system with a spring constant of 100 N/m is oscillating vertically with an amplitude of 0.200 m. When the mass reaches the top of its oscillation, it is hit with an object that gives it a downward velocity of 1.00 m/s. Determine the total mechanical energy before being hit, the speed at the equilibrium point after being hit, and the new amplitude of the oscillation. Solution: Step 1: Find the total mechanical energy At the top of the oscillation, the potential energy of the spring is: \(U = \frac12 kx^2 = \frac12 (100\,\mathrm{N/m})(0.200\,\mathrm{m})^2 = 2.00\,\mathrm{J}\) The kinetic energy at the top of the oscillation is zero: \(K = 0\,\mathrm{J}\) After being hit, the new initial kinetic energy is: \(K = \frac{1}{2}mv^2 = \frac{1}{2}(1.00\,\mathrm{kg})(1.00\,\mathrm{m/s})^2 = 0.500\,\mathrm{J}\) The total mechanical energy at the top after being hit is: \(E_{top} = K + U = 0.500\,\mathrm{J} + 2.00\,\mathrm{J} = 2.50\,\mathrm{J}\) Step 2: Calculate the speed at the equilibrium point At the equilibrium point, the potential energy is zero. Using conservation of mechanical energy: \(E_{top} = K_{eq}\) ⟹ \(2.50\,\mathrm{J} = \frac{1}{2}(1.00\,\mathrm{kg})v_{eq}^2\) Solving for \(v_{eq}\), we get: \(v_{eq} = \sqrt{\frac{2(2.50\,\mathrm{J})}{1.00\,\mathrm{kg}}} = 2.24\,\mathrm{m/s}\) Step 3: Determine the new amplitude At the new maximum position, the mechanical energy is: \(E_{new\,max} = \frac12 kx_{new}^2\) Using conservation of energy: \(E_{top} = E_{new\,max}\) ⟹ \(2.50\,\mathrm{J} = \frac12 (100\,\mathrm{N/m})x_{new}^2\) Solving for \(x_{new}\), we get: \(x_{new} = \sqrt{\frac{2(2.50\,\mathrm{J})}{100\,\mathrm{N/m}}} = 0.316\,\mathrm{m}\) Answers: 1. The total mechanical energy before being hit is 2.00 J. 2. The speed at the equilibrium point after being hit is 2.24 m/s. 3. The new amplitude of the oscillation is 0.316 m.

Step-by-step solution

Find the total mechanical energy

To find the total mechanical energy, we first need to determine the potential energy and kinetic energy of the system at the top of its oscillation (before being hit). Potential energy can be calculated using the formula: \(U = \frac12 kx^2\) where \(U\) is the potential energy of the spring, \(k = 100 \mathrm{~N/m}\) is the spring constant, and \(x = 0.200 \mathrm{~m}\) is the amplitude. Similarly, the kinetic energy at the top of the oscillation is zero because the mass is momentarily at rest. After being hit, the mass has a downward velocity of \(1.00 \mathrm{~m/s}\). Consequently, the new initial kinetic energy can be calculated using the formula: \(K = \frac{1}{2}mv^2\) where \(K\) is the kinetic energy, \(m = 1.00 \mathrm{~kg}\) is the mass, and \(v = 1.00 \mathrm{~m/s}\) is the velocity after being hit. Now we can calculate the total mechanical energy (potential + kinetic energy).

Calculate the speed at the equilibrium point

At the equilibrium point, the potential energy of the spring is zero. Using the conservation of mechanical energy principle, the total mechanical energy remains constant: \(E_{top} = E_{equilibrium}\) \(E_{top}\) represents the total mechanical energy at the top, and \(E_{equilibrium}\) represents the total mechanical energy at the equilibrium position. We can find the speed at the equilibrium (\(v_{eq}\)) using this equation by first setting \(E_{top} = K_{top} + U_{top}\) and \(E_{equilibrium} = K_{eq}\), then solving for \(v_{eq}\).

Determine the new amplitude

To find the new amplitude (\(x_{new}\)), we can use the conservation of energy principle again: \(E_{top} = E_{new\,max}\) Where \(E_{new\,max}\) represents the total mechanical energy at the new maximum position. At this point, all the mechanical energy will be stored as potential energy in the spring: \(E_{new\,max} = U_{new\,max} = \frac12 kx_{new}^2\) Solving for \(x_{new}\) will provide the new amplitude.

Key concepts

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction. This restoring force can come from a variety of sources, such as the tension in a pendulum or the compression and extension in a spring.

What's fascinating about SHM is that it follows a sine or cosine wave over time when plotted on a graph. If you picture a mass oscillating on a spring, it passes through a central 'equilibrium point' where the force is zero. On either side of this point, the spring exerts a force to pull the mass back towards it, creating the oscillation.

For instance, in the solved exercise, before the mass was hit and began moving, it was exhibiting SHM. The amplitude of the motion—the maximum distance from the equilibrium position—is an essential characteristic of SHM. It helps determine the energy involved in the oscillation.

Conservation of Mechanical Energy

One of the cornerstones of classical physics is the Conservation of Mechanical Energy principle. It states that in a closed system with no external forces, the total mechanical energy remains constant. In other words, energy can neither be created nor destroyed; it can only be transformed from one form to another.

In SHM, like the example where a mass is attached to a spring, the mechanical energy is continuously converted between potential and kinetic energy, but the total amount remains constant unless an external force does work on the system.

When the mass in our exercise is hit and given additional kinetic energy, this energy is added to the system, increasing the total mechanical energy. No energy is lost or gained thereafter; it just changes form as the mass continues to oscillate.

Spring Potential Energy

The potential energy in a spring, often termed as spring potential energy, is a type of elastic potential energy. It's the energy stored when we stretch or compress a spring. The formula for spring potential energy is:\[U = \frac12 kx^2\]

Here, \(U\) is the potential energy, \(k\) is the spring constant (a measure of the spring's stiffness), and \(x\) is the displacement from the equilibrium position.

The exercise in question illustrates how to calculate the potential energy stored in a spring when a mass hangs from it and is displaced. Initially, when the mass is at the top of its oscillation, all the energy is stored as potential energy. As the mass moves, this energy will convert into kinetic energy and vice versa.

Kinetic Energy

Kinetic energy is the energy of motion. Any object that is moving has kinetic energy, and this energy depends on the mass of the object and its velocity. The kinetic energy (\(K\)) can be calculated with the formula:\[K = \frac{1}{2}mv^2\]

Here, \(m\) represents the mass of the object and \(v\) its velocity. In the context of our exercise, kinetic energy is zero at the highest and lowest points of the oscillation, where the velocity is momentarily zero. However, as soon as the mass is hit and gains speed, its kinetic energy increases.

This kinetic energy reaches its peak when the spring passes through the equilibrium point since this is when the mass has its maximum velocity. Following the conservation of energy principle, at the equilibrium point, all of the system's mechanical energy is kinetic.