From f3834ca40d75b8b1913fad28cba0b108ac446bcf Mon Sep 17 00:00:00 2001 From: Mars Ultor Date: Tue, 28 Jan 2025 20:29:39 -0600 Subject: [PATCH] i'm going to fail this test --- content/physics/harmonic-motion.md | 197 ++++++++++++++++++++++++++++- 1 file changed, 196 insertions(+), 1 deletion(-) diff --git a/content/physics/harmonic-motion.md b/content/physics/harmonic-motion.md index 9e6347a..b7c6899 100644 --- a/content/physics/harmonic-motion.md +++ b/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**. +