292 lines
7.8 KiB
Markdown
292 lines
7.8 KiB
Markdown
---
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title: Less Simple Harmonic Motion
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date: 2025-01-09
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---
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# Simple harmonic motion - Spring oscillations
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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.
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$$
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F=\pm kx
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$$
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Terms, I guess:
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- the distance from the equilibrium point is *displacement*
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- Peak displacement is called the *amplitude*
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- one *cycle* is the distance covered between the trough of the position curve over time and the peak of the same curve
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- the *period* $T$ is the time required to complete one cycle's worth of motion
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- the *frequency* $f$ is the number of complete cycles per second.
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$$
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f=\frac{1}{T} \text{ and } T=\frac{1}{f}
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$$
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# Energy in Simple Harmonic Motion
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Please just go learn Lagrangian Mechanics already. The energy approach is obviously the most useful of them all. Elastic Potential energy:
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$$
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|\int{F\mathrm{d}x}|=\int{kx \mathrm{d}x}=\tfrac{1}{2}kx^2
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$$
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The total energy: $E=\textbf{KE}+\textbf{PE}$
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$$
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E=\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2
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$$
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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 $\textbf{KE}=0$.
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$$
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\tfrac{1}{2}m0^2+\tfrac{1}{2}kA^2
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$$
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At equilibrium, velocity and therefore kinetic energy are the greatest, with $\textbf{PE}=0$:
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$$
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E=\tfrac{1}{2}m(v_{\text{max}})^2
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$$
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Since energy is always conserved in SHM:
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$$
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\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2
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$$
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Solving the system of equations for $v$:
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$$
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v^2 = \frac{k}{m}A^2(1-\frac{x^2}{A^2})
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$$
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Given $mv^2=kA^2$:
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$$
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v_{\text{max}}=\sqrt{\frac{k}{m}}A
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$$
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With another step of substitution:
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$$
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v=\pm v_{\text{max}}\sqrt{1-\frac{x^2}{A^2}}
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$$
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# The Period and the Sinusoidal Nature of SHM
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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 ,$A$.
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$$
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T=2\pi\sqrt{\tfrac{m}{k}}
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$$
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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 $k\propto F$
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$$
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f=\frac{1}{t}=\frac{1}{2\pi}\sqrt{\frac{k}{m}}
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$$
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Here's the expanded derivation with the characteristic equation and solution in a markdown format with LaTeX code blocks:
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---
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# Derivation of the Mass-Spring SHM Equation Using Lagrangian Mechanics
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We start with the equation of motion derived from the Lagrangian approach:
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$$
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m \ddot{x} + kx = 0
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$$
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## Step 1: Rewrite as a Second-Order ODE
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$$
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\ddot{x} + \frac{k}{m}x = 0
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$$
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Let:
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$$
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\omega^2 = \frac{k}{m}
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$$
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Thus, the equation becomes:
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$$
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\ddot{x} + \omega^2 x = 0
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$$
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---
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## Step 2: Characteristic Equation
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To solve this second-order differential equation, assume a solution of the form:
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$$
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x(t) = e^{\lambda t}
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$$
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Substitute $ $x(t) = e^{\lambda t}$ into the differential equation:
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$$
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\lambda^2 e^{\lambda t} + \omega^2 e^{\lambda t} = 0
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$$
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Factoring out $ $e^{\lambda t}$ (which is never zero):
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$$
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\lambda^2 + \omega^2 = 0
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$$
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The characteristic equation is:
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$$
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\lambda^2 + \omega^2 = 0
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$$
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---
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## Step 3: Solve for $ $\lambda$
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Rearranging:
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$$
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\lambda^2 = -\omega^2
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$$
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Taking the square root:
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$$
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\lambda = \pm i\omega
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$$
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Thus, the general solution for $ $x(t)$ is a linear combination of exponential functions:
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$$
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x(t) = C_1 e^{i\omega t} + C_2 e^{-i\omega t}
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$$
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---
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## Step 4: Convert to Real Solution
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Using Euler's formula:
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$$
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e^{i\omega t} = \cos(\omega t) + i\sin(\omega t)
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$$
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$$
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e^{-i\omega t} = \cos(\omega t) - i\sin(\omega t)
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$$
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The general solution becomes:
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$$
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x(t) = C_1 (\cos(\omega t) + i\sin(\omega t)) + C_2 (\cos(\omega t) - i\sin(\omega t))
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$$
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Grouping real and imaginary parts, let:
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$$
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A = C_1 + C_2, \quad B = i(C_1 - C_2)
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$$
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The solution can then be written as:
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$$
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x(t) = A \cos(\omega t) + B \sin(\omega t)
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$$
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---
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## Step 5: Final General Solution
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Alternatively, write the solution in amplitude-phase form:
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$$
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x(t) = C \cos(\omega t + \phi)
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$$
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where:
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- $C = \sqrt{A^2 + B^2}$ (amplitude),
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- $\phi = \tan^{-1}\left(\frac{B}{A}\right)$ (phase angle).
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This is the general solution for the simple harmonic motion of a mass-spring system.
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# Standing Waves
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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:
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- **Both Ends Closed** (e.g., a string fixed at both ends or a closed pipe)
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- **One End Closed, One End Open** (e.g., wind instruments like clarinets)
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- **Both Ends Open** (e.g., wind instruments like flutes).
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## **Both Ends Closed (Fixed String or Closed Pipe)**
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- **Boundary conditions:** Nodes at both ends (zero displacement).
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- **Wave pattern:** The tube must fit an integer number of **half-wavelengths**.
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### **Resonance Condition:**
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$$
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L = n \frac{\lambda}{2}, \quad n = 1, 2, 3, \dots
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$$
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### **Frequency Formula:**
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Using $ v = f \lambda $, solving for $ f $:
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$$
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f_n = \frac{n v}{2L}, \quad n = 1, 2, 3, \dots
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$$
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### **How to Calculate $ n $:**
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Since $ n $ counts both nodes and antinodes:
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$$
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n = \text{total antinodes}
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$$
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---
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## **One End Closed, One End Open (Pipe with One Closed End)**
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- **Boundary conditions:** A node at the closed end (zero displacement) and an antinode at the open end (maximum displacement).
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- **Wave pattern:** The tube must fit an **odd number of quarter-wavelengths**.
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### **Resonance Condition:**
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$$
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L = n \frac{\lambda}{4}, \quad n = 1, 3, 5, 2k-1, \dots
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$$
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### **Frequency Formula:**
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Using $ v = f \lambda $, solving for $ f $:
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$$
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f_n = \frac{n v}{4L}, \quad n = 1, 3, 5, 2k-1 , \dots
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$$
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### **How to Calculate $ n $:**
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Since $ n $ counts both nodes and antinodes:
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$$
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n = \text{total nodes} + \text{total antinodes}
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$$
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---
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## **Both Ends Open (Pipe Open at Both Ends)**
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- **Boundary conditions:** Antinodes at both ends (maximum displacement).
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- **Wave pattern:** The tube must fit an integer number of **half-wavelengths**, similar to the **both ends closed** case.
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### **Resonance Condition:**
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$$
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L = n \frac{\lambda}{2}, \quad n = 1, 2, 3, \dots
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$$
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### **Frequency Formula:**
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Using $ v = f \lambda$, solving for $ f $:
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$$
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f_n = \frac{n v}{2L}, \quad n = 1, 2, 3, \dots
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$$
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### **How to Calculate $ n $:**
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Since $ n $ counts both nodes and antinodes:
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$$
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n = \text{total nodes} = \text{total antinodes} -1
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$$
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---
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## **Summary Table: Resonance Conditions and Harmonics Calculation**
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| Condition | Node-Antinode Pattern | Wavelength Relation | Frequency Formula | Harmonic Number Calculation |
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|-----------------------------|------------------------------|--------------------------------------------|----------------------------|-----------------------------|
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| **Both Ends Closed** | Node at both ends | $L = n \frac{\lambda}{2}$ | $f_n = \frac{n v}{2L}$ | $n = \text{antinodes}$ |
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| **One End Closed, One Open**| Node at closed, antinode at open | $L = n \frac{\lambda}{4}$ | $f_n = \frac{n v}{4L}$ | $n = \text{nodes} + \text{antinodes}$ |
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| **Both Ends Open** | Antinode at both ends | $L = n \frac{\lambda}{2}$ | $f_n = \frac{n v}{2L}$ | $n = \text{antinodes}$ |
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Now the formulas **remain internally consistent with a fixed definition of $ n $** across all cases.
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