---
title: Less Simple Harmonic Motion
date: 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 cases.