Question

A 2.00 -kg mass attached to a spring is displaced \(8.00 \mathrm{~cm}\) from the equilibrium position. It is released and then oscillates with a frequency of \(4.00 \mathrm{~Hz}\) a) What is the energy of the motion when the mass passes through the equilibrium position? b) What is the speed of the mass when it is \(2.00 \mathrm{~cm}\) from the equilibrium position?

Short answer

Answer: To find the speed of the mass when it is 2.00 cm from the equilibrium position, follow these steps: 1. Calculate the angular frequency (ω) using the given frequency (4.00 Hz): ω = 2πf 2. Solve for the spring constant (k) using the angular frequency and mass (0.500 kg) formulas: k = mω² 3. Calculate the energy (E) of the motion at the equilibrium position using the amplitude (8.00 cm) and spring constant: E = (1/2) kA² 4. Find the potential energy (U) when the mass is 2.00 cm from the equilibrium position: U = (1/2) kx² 5. Calculate the kinetic energy (K) at that position: K = E - U 6. Find the speed (v) of the mass at that position using the kinetic energy formula (K = (1/2) mv²) and solving for v. After following the steps and solving for v, you will find the speed of the mass when it is 2.00 cm from the equilibrium position.

Step-by-step solution

1. Calculate the angular frequency

To calculate the energy and speed at different positions, we first need to find the angular frequency (ω) of the oscillation. We can find it using the given frequency (f): ω = 2πf Where f = 4.00 Hz.

2. Calculate the energy of motion

When the mass is at the equilibrium position, all the potential energy is converted into kinetic energy. So, the total energy (E) of the system at that time equals kinetic energy (K): E = K We can find the energy at the equilibrium position using the formula for the potential energy when it is displaced 8.00 cm (maximum displacement or amplitude, A) from the equilibrium position: E = (1/2) kA² We also have the formula for angular frequency in terms of mass (m) and spring constant (k) as: ω = sqrt(k/m) Now, we can solve for k and substitute it into the energy equation.

3. Solve for k and calculate the energy

First, find k using the angular frequency and mass formulas: ω² = k/m We know ω and m so, we can solve for k: k = mω² Now, substitute the k value in the energy formula: E = (1/2) kA² With the given A, we can find the energy (E) of the motion when the mass passes through the equilibrium position.

4. Find the speed when the mass is 2.00 cm from the equilibrium position

At any moment, the energy of the system can be expressed as the sum of potential energy (U) and kinetic energy (K): E = U + K When the mass is 2.00 cm from the equilibrium position, we can find its potential energy using the formula: U = (1/2) kx² Where x = 2.00 cm is the current displacement of the mass. Now, we can find the kinetic energy (K) at that position: K = E - U Then, we can find the speed (v) of the mass at that position using the kinetic energy formula: K = (1/2) mv² Finally, solving for v will give us the speed of the mass at 2.00 cm from the equilibrium position.

Key concepts

Angular Frequency

Angular frequency, often represented by the Greek letter \(\omega\), is a fundamental parameter in oscillatory motion physics, particularly in the study of simple harmonic motion (SHM). It measures how quickly an object oscillates about the equilibrium position and is directly related to the frequency \(f\) of oscillation.
In the context of a mass attached to a spring, \(\omega\) can be calculated using the relationship \(\omega = 2\pi f\). For example, if the frequency is given as 4.00 Hz, the angular frequency would be \(\omega = 2\pi \times 4.00 \mathrm{Hz}\). This calculation is crucial because it allows us to determine other important quantities, such as the total energy in the system and the speed of the oscillating mass at different positions.
Understanding \(\omega\) is essential as it also relates the spring constant \(k\) and the mass \(m\) of the object in SHM through the formula \(\omega = \sqrt{k/m}\). With the mass known, calculating \(k\) becomes possible, which then enables further analysis of the motion.

Kinetic and Potential Energy

Kinetic and potential energy play integral roles in the analysis of oscillatory systems. In the context of SHM, when an object passes through the equilibrium position, its potential energy is zero, and all the energy of the system is kinetic.
Kinetic energy \(K\) is given by \(K = (1/2)mv^2\), where \(m\) is the mass and \(v\) is the velocity of the object. On the other hand, potential energy \(U\) in a spring system is given by \(U = (1/2)kx^2\), where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position.
At the maximum displacement (amplitude), all the system's energy is potential, and as the mass moves towards the equilibrium position, this potential energy is converted into kinetic energy. The total mechanical energy \(E\) of the system is conserved, meaning it remains constant, being interchangeably transformed between kinetic and potential forms during the oscillation.

Simple Harmonic Motion

Simple harmonic motion (SHM) is characterized by a restoring force that is proportional to the displacement and acts in the opposite direction. For a mass-spring system, Hooke's Law describes this restoring force as \(F = -kx\), which leads to SHM if no other forces are acting on the system.
A system in SHM will have a sinusoidal motion with a period that depends on the mass of the object and the spring constant. The beauty of SHM is that regardless of the amplitude of the motion, the period remains constant. This highlights that SHM depends solely on the intrinsic properties (mass and spring constant) of the system and not on the amplitude of the oscillations.
A typical SHM problem involves calculating the period, the maximum velocity, or the energies at various points. These calculations often hinge on understanding the relationships between period, frequency, angular frequency, and the energies involved in the motion.

Spring Constant Calculation

The spring constant \(k\), a measure of a spring's stiffness, is integral to problems involving SHM. Higher values of \(k\) represent stiffer springs which exert greater restoring forces for the same displacement. The spring constant is calculated by rearranging the formula for angular frequency \(\omega = \sqrt{k/m}\). By squaring both sides, we get \(\omega^2 = k/m\), which allows us to solve for \(k\) when \(\omega\) and \(m\) are known.
For example, in the given exercise, once we know the angular frequency and the mass of the object attached to the spring, we can calculate \(k\) as \(k = m\omega^2\). This value helps us find the total energy at equilibrium and also aids in calculating the speed of the mass at different displacements from equilibrium by allowing us to determine both kinetic and potential energies.