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content/physics/harmonic-motion.md

@@ -62,7 +62,7 @@ v^2 = \frac{k}{m}A^2(1-\frac{x^2}{A^2})
 $$
 Given $mv^2=kA^2$:
 $$
-v_max=\sqrt{\frac{k}{m}}A 
+v_{\text{max}}=\sqrt{\frac{k}{m}}A 
 $$
 
 With another step of substitution:
@@ -85,3 +85,198 @@ $$
 f=\frac{1}{t}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}
 $$
 
+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 occur when waves reflect and interfere in a confined space, forming resonance at specific frequencies. The resonance conditions depend on the boundary conditions: both ends closed, one end closed and the other open, or both ends open.
+
+---
+
+## **Both Ends Closed**
+
+When both ends of a tube or string are closed, **nodes** form at both ends because there is no displacement at the boundaries. The wave pattern consists of an integer number of **half-wavelengths** fitting inside the length  $L$.
+
+### **Resonance Condition:**
+$$
+L = n \frac{\lambda}{2}, \quad n = 1, 2, 3, \dots
+$$
+
+### **Resonant Frequencies:**
+$$
+f_n = \frac{n v}{2L}, \quad n = 1, 2, 3, \dots
+$$
+
+### **How to Calculate $n$:**
+For this case, the harmonic number $n$ is simply the **number of half-wavelengths in the tube**:
+$$
+n = \text{number of nodes} - 1 = \text{number of antinodes}
+$$
+
+---
+
+## **One End Closed, One End Open**
+
+When one end is closed, a **node** forms at the closed end (no displacement), and an **antinode** forms at the open end (maximum displacement). Only **odd harmonics** fit in this setup because the length must accommodate an odd number of quarter-wavelengths.
+
+### **Resonance Condition:**
+$$
+L = (2n-1) \frac{\lambda}{4}, \quad n = 1, 2, 3, \dots
+$$
+
+### **Resonant Frequencies:**
+$$
+f_n = \frac{(2n-1) v}{4L}, \quad n = 1, 2, 3, \dots
+$$
+
+### **How to Calculate $n$:**
+For this case, $n$ is determined by the **sum of nodes and antinodes**:
+$$
+n = \text{number of nodes} + \text{number of antinodes}
+$$
+Since there is always **one  $n = 1, 2, 3, \dots$.
+
+---
+
+## **Both Ends Open**
+
+When both ends are open, **antinodes** form at both ends because displacement is maximum at the boundaries. The standing wave pattern is identical to the **both ends closed** case.
+
+### **Resonance Condition:**
+$$
+L = n \frac{\lambda}{2}, \quad n = 1, 2, 3, \dots
+$$
+
+### **Resonant Frequencies:**
+$$
+f_n = \frac{n v}{2L}, \quad n = 1, 2, 3, \dots
+$$
+
+### **How to Calculate $                                                $ n $:**
+Since an antinode exists at each end, the harmonic number $                                                $ n $ corresponds to the number of **antinodes**:
+$$
+n = \text{number of antinodes}
+$$
+
+---
+
+## **Summary Table: Resonance Conditions and Harmonics Calculation**
+
+| Condition                  | Node-Antinode Pattern         | Wavelength Relation                        | Resonant Frequency       | Harmonic Number Calculation |
+|----------------------------|------------------------------|--------------------------------------------|--------------------------|-----------------------------|
+| **Both Ends Closed**       | Node at both ends             |  $L = n \frac{\lambda}{2}$               | $f_n = \frac{n v}{2L}$ | $n = \text{nodes} - 1 = \text{antinodes}$ |
+| **One End Closed, One Open** | Node at closed, antinode at open |        $L = (2n-1) \frac{\lambda}{4}$          | $f_n = \frac{(2n-1)v}{4L}$ | $n = \text{nodes} + \text{antinodes}$ |
+| **Both Ends Open**         | Antinode at both ends         |  $L = n \frac{\lambda}{2}$               | $f_n = \frac{n v}{2L}$ | $n = \text{antinodes}$ |
+
+Each case follows the fundamental rule that resonance occurs when the physical constraints match a standing wave pattern, leading to **natural frequencies of oscillation**.
+