warehouse/content/physics/harmonic-motion.md

title, date

title	date
Less Simple Harmonic Motion	2025-01-09

Simple harmonic motion - Spring oscillations

When an object vibrates or oscillates back and forth, over the same path, each oscillation taking the same amount of time, the motion is periodic. The simplest form of periodic motion closely resemble this system, we will examine ... blah blah blah. Any spring at natural length is defined to be at its equilibrium position. If additional energy is imparted upon the system, either compressing or stretching said spring, a restoring force will occur to restore the initial position.

F=\pm kx

Terms, I guess:

• the distance from the equilibrium point is displacement
• Peak displacement is called the amplitude
• one cycle is the distance covered between the trough of the position curve over time and the peak of the same curve
• the period T is the time required to complete one cycle's worth of motion
• the frequency f is the number of complete cycles per second.

f=\frac{1}{T} \text{ and } T=\frac{1}{f}

Energy in Simple Harmonic Motion

Please just go learn Lagrangian Mechanics already. The energy approach is obviously the most useful of them all. Elastic Potential energy:

|\int{F\mathrm{d}x}|=\int{kx \mathrm{d}x}=\tfrac{1}{2}kx^2 

The total energy: E=\textbf{KE}+\textbf{PE}

E=\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 

We all know what x and v are. SHM can only occur if friction and other external forces are approximately net zero. A is defined as the absolute peak distance from equilibrium. When x=\pm A, the \textbf{KE}=0.

\tfrac{1}{2}m0^2+\tfrac{1}{2}kA^2 

At equilibrium, velocity and therefore kinetic energy are the greatest, with \textbf{PE}=0:

E=\tfrac{1}{2}m(v_{\text{max}})^2 

Since energy is always conserved in SHM:

\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2 

Solving the system of equations for v:

v^2 = \frac{k}{m}A^2(1-\frac{x^2}{A^2})

Given mv^2=kA^2:

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

With another step of substitution:

v=\pm v_{\text{max}}\sqrt{1-\frac{x^2}{A^2}}

The Period and the Sinusoidal Nature of SHM

The period of harmonic motion depends on the stiffness of the spring and the mass m in question. It does not, however, depend on the amplitude ,$A$.

T=2\pi\sqrt{\tfrac{m}{k}}

Larger mass means more rotational inertial (we haven't gotten to this yet), so that part of it makes sense. It is also important that k\propto F

f=\frac{1}{t}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Here's the expanded derivation with the characteristic equation and solution in a markdown format with LaTeX code blocks:

---

Derivation of the Mass-Spring SHM Equation Using Lagrangian Mechanics

We start with the equation of motion derived from the Lagrangian approach:

m \ddot{x} + kx = 0

Step 1: Rewrite as a Second-Order ODE

\ddot{x} + \frac{k}{m}x = 0

Let:

\omega^2 = \frac{k}{m}

Thus, the equation becomes:

\ddot{x} + \omega^2 x = 0

---

Step 2: Characteristic Equation

To solve this second-order differential equation, assume a solution of the form:

x(t) = e^{\lambda t}

Substitute $ x(t) = e^{\lambda t} into the differential equation:

\lambda^2 e^{\lambda t} + \omega^2 e^{\lambda t} = 0

Factoring out $ e^{\lambda t} (which is never zero):

\lambda^2 + \omega^2 = 0

The characteristic equation is:

\lambda^2 + \omega^2 = 0

---

Step 3: Solve for $ \lambda

Rearranging:

\lambda^2 = -\omega^2

Taking the square root:

\lambda = \pm i\omega

Thus, the general solution for $ x(t) is a linear combination of exponential functions:

x(t) = C_1 e^{i\omega t} + C_2 e^{-i\omega t}

---

Step 4: Convert to Real Solution

Using Euler's formula:

e^{i\omega t} = \cos(\omega t) + i\sin(\omega t)

e^{-i\omega t} = \cos(\omega t) - i\sin(\omega t)

The general solution becomes:

x(t) = C_1 (\cos(\omega t) + i\sin(\omega t)) + C_2 (\cos(\omega t) - i\sin(\omega t))

Grouping real and imaginary parts, let:

A = C_1 + C_2, \quad B = i(C_1 - C_2)

The solution can then be written as:

x(t) = A \cos(\omega t) + B \sin(\omega t)

---

Step 5: Final General Solution

Alternatively, write the solution in amplitude-phase form:

x(t) = C \cos(\omega t + \phi)

where:

• C = \sqrt{A^2 + B^2} (amplitude),
• \phi = \tan^{-1}\left(\frac{B}{A}\right) (phase angle).

This is the general solution for the simple harmonic motion of a mass-spring system.

Standing Waves

Standing waves form in a confined space due to wave interference, leading to resonant frequencies where the wave reinforces itself. The resonance conditions differ based on boundary conditions:

• Both Ends Closed (e.g., a string fixed at both ends or a closed pipe)
• One End Closed, One End Open (e.g., wind instruments like clarinets)
• Both Ends Open (e.g., wind instruments like flutes).

Both Ends Closed (Fixed String or Closed Pipe)

• Boundary conditions: Nodes at both ends (zero displacement).
• Wave pattern: The tube must fit an integer number of half-wavelengths.

Resonance Condition:

L = n \frac{\lambda}{2}, \quad n = 1, 2, 3, \dots

Frequency Formula:

Using v = f \lambda, solving for f:

f_n = \frac{n v}{2L}, \quad n = 1, 2, 3, \dots

How to Calculate n:

Since n counts both nodes and antinodes:

n = \text{total antinodes}

---

One End Closed, One End Open (Pipe with One Closed End)

• Boundary conditions: A node at the closed end (zero displacement) and an antinode at the open end (maximum displacement).
• Wave pattern: The tube must fit an odd number of quarter-wavelengths.

Resonance Condition:

L = n \frac{\lambda}{4}, \quad n = 1, 3, 5, 2k-1, \dots

Frequency Formula:

Using v = f \lambda, solving for f:

f_n = \frac{n v}{4L}, \quad n = 1, 3, 5, 2k-1 , \dots

How to Calculate n:

Since n counts both nodes and antinodes:

n = \text{total nodes} + \text{total antinodes}

---

Both Ends Open (Pipe Open at Both Ends)

• Boundary conditions: Antinodes at both ends (maximum displacement).
• Wave pattern: The tube must fit an integer number of half-wavelengths, similar to the both ends closed case.

Resonance Condition:

L = n \frac{\lambda}{2}, \quad n = 1, 2, 3, \dots

Frequency Formula:

Using v = f \lambda, solving for f:

f_n = \frac{n v}{2L}, \quad n = 1, 2, 3, \dots

How to Calculate n:

Since n counts both nodes and antinodes:

n =  \text{total nodes} = \text{total antinodes} -1

---

Summary Table: Resonance Conditions and Harmonics Calculation

Condition	Node-Antinode Pattern	Wavelength Relation	Frequency Formula	Harmonic Number Calculation
Both Ends Closed	Node at both ends	L = n \frac{\lambda}{2}	f_n = \frac{n v}{2L}	n = \text{antinodes}
One End Closed, One Open	Node at closed, antinode at open	L = n \frac{\lambda}{4}	f_n = \frac{n v}{4L}	n = \text{nodes} + \text{antinodes}
Both Ends Open	Antinode at both ends	L = n \frac{\lambda}{2}	f_n = \frac{n v}{2L}	n = \text{antinodes}

Now the formulas remain internally consistent with a fixed definition of $ n $ across all case.