Energy types in oscillating systems

Oscillating systems, such as a mass attached to a spring, undergo simple harmonic motion when displaced from their equilibrium position and released. In these systems, energy exists in two main forms: kinetic energy, which is the energy of motion, and elastic potential energy, which is stored in the deformed spring.

This energy setup allows the system to oscillate back and forth. As the mass moves, energy continuously interchanges between these two forms, but the total energy remains constant if no external forces like friction are present.

Key energy forms:

• Kinetic energy (E_k) - The energy due to the velocity of the moving mass
• Elastic potential energy (E_e) - The energy stored when the spring is stretched or compressed from its natural length

Continuous interchange between kinetic and elastic potential energy

In an oscillating system, energy constantly shifts between kinetic and elastic potential forms as the mass moves through its cycle. This interchange drives the back-and-forth motion without any net loss of energy in an ideal system.

At any point, the sum of kinetic and elastic potential energy equals the total energy. This relationship allows us to understand how the system's behavior changes during oscillation.

How energy interchanges during oscillation:

• When the spring is stretched or compressed, elastic potential energy increases while kinetic energy decreases as the mass slows down.
• As the mass passes through equilibrium, kinetic energy increases while elastic potential energy decreases.
• This back-and-forth conversion repeats with each cycle, maintaining the oscillation.

Points of maximum potential and kinetic energy during oscillation

Energy distribution varies at different positions in the oscillation cycle. Displacement extremes and the equilibrium position represent key points where one form of energy reaches its maximum while the other is zero.

Maximum elastic potential energy

• Occurs at displacement extremes, where the mass is at its maximum distance from equilibrium (amplitude).
• At these points, velocity is zero, so kinetic energy is zero.
• All energy is stored as elastic potential energy in the spring.

Maximum kinetic energy

• Occurs at the equilibrium position, where displacement is zero.
• At this point, elastic potential energy is zero because the spring is at its natural length.
• All energy exists as kinetic energy, with the mass moving at its maximum velocity.

Conservation of total energy as a constant value

The total energy (E_total) in the system remains constant throughout the oscillation, following the principle of energy conservation. This means E_total = E_k + E_e = constant at every point in the cycle.

This conservation allows us to predict the system's behavior, such as how velocity changes with position or how far the mass will travel.

Deriving the relationship between total energy and amplitude

The total energy of the system can be expressed in terms of the spring constant (k) and amplitude (A), which is the maximum displacement from equilibrium.

At displacement extremes, where kinetic energy is zero, all energy is elastic potential energy. The formula for elastic potential energy is $\frac{1}{2}k{x}^2$, where x is displacement. At amplitude, x = A, so total energy equals $\frac{1}{2}k{A}^2$.

Formula for total energy:

$E_{\text{total}}=\frac{1}{2}k{A}^2$

Where:

• E_total = Total energy (joules, J)
• k = Spring constant (newtons per meter, N/m)
• A = Amplitude (meters, m)

This relationship shows that total energy depends on how stiff the spring is and how far it is initially displaced.

Deriving the relationship for maximum velocity

Maximum velocity occurs at the equilibrium position, where all energy is kinetic. The formula for kinetic energy is $\frac{1}{2}m{v}^2$, where m is mass and v is velocity.

Since total energy is constant and equals $\frac{1}{2}k{A}^2$, at equilibrium: $\frac{1}{2}m{v}_{\text{max}}^2=\frac{1}{2}k{A}^2$.

Solving for v_max gives $v_{\text{max}}=A\sqrt{\frac{k}{m}}$.

Formula for maximum velocity:

$v_{\text{max}}=A\sqrt{\frac{k}{m}}$

Where:

• v_max = Maximum velocity (meters per second, m/s)
• A = Amplitude (meters, m)
• k = Spring constant (newtons per meter, N/m)
• m = Mass (kilograms, kg)

This shows that maximum velocity increases with larger amplitude or stiffer springs but decreases with greater mass.

Applying energy conservation to predict velocities and positions

Energy conservation equations can predict velocities at specific positions or positions where certain velocities occur. Use $E_{\text{total}}=E_k+E_e=\frac{1}{2}m{v}^2+\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2$.

To find velocity at a position x, rearrange to $v=\sqrt{\frac{k}{m}(A^2-x^2)}$.

To find position for a given velocity, rearrange accordingly.

Worked example - Predicting velocity at a specific position

A 0.5 kg mass on a spring with k = 200 N/m oscillates with amplitude 0.1 m. Calculate the velocity when displacement is 0.05 m from equilibrium.

Step 1: Identify the values

• Mass (m) = 0.5 kg
• Spring constant (k) = 200 N/m
• Amplitude (A) = 0.1 m
• Displacement (x) = 0.05 m

Step 2: Apply the formula

Use $v=\sqrt{\frac{k}{m}(A^2-x^2)}$

Step 3: Substitution and calculation

$v=\sqrt{\frac{200}{0.5}({0.1}^2-{0.05}^2)}=\sqrt{400(0.01-0.0025)}=\sqrt{400\times 0.0075}=\sqrt{3}\approx 1.73\text{ m/s}$

The velocity is approximately 1.73 m/s at 0.05 m displacement.

Worked example - Predicting position for a given velocity

In the same system (m = 0.5 kg, k = 200 N/m, A = 0.1 m), find the displacement where velocity is 1.0 m/s.

Step 1: Rearrange the formula

From $v=\sqrt{\frac{k}{m}(A^2-x^2)}$

rearrange to $x=\sqrt{A^2-\frac{m{v}^2}{k}}$

Step 2: Substitution and calculation

$x=\sqrt{{0.1}^2-\frac{0.5\times {1.0}^2}{200}}=\sqrt{0.01-\frac{0.5}{200}}=\sqrt{0.01-0.0025}=\sqrt{0.0075}\approx 0.087\text{ m}$

The displacement is approximately 0.087 m when velocity is 1.0 m/s.