warehouse/content/physics/harmonic-motion.md

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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:

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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

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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

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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}

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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)

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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}

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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}-1

The above is true only if you cound the starting node of the fixture of the air column. You do not subtract by 1 otherwise.

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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

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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.

Pendulum formulas

Here are the main pendulum formulas in general physics:

1. Period of a simple pendulum (small angle approximation):

   
   T = 2\pi \sqrt{\frac{L}{g}}
   

   Where:

   • T = period (time for one full swing)
   • L = length of the pendulum
   • g = acceleration due to gravity

2. Frequency of a simple pendulum:

   
   f = \frac{1}{T} = \frac{1}{2\pi} \sqrt{\frac{g}{L}}
   

   Where:

   • f = frequency (number of oscillations per second)

3. Angular frequency of a simple pendulum: $ \omega = \sqrt{\frac{g}{L}} $ Where:

   • \omega = angular frequency (rad/s)

4. Displacement of a simple pendulum (for small angles):

   
   \theta(t) = \theta_0 \cos(\omega t)
   

   Where:

   • \theta(t) = angular displacement at time (t)
   • \theta_0 = maximum displacement (amplitude)

5. Energy of a simple pendulum:

   • Potential Energy at maximum displacement:

     
     PE = mgh
     

     Where:

     • m = mass of the pendulum bob
     • h = height relative to the lowest point

   • Kinetic Energy at the equilibrium point:

     
     KE = \frac{1}{2}mv^2
     

6. Damped pendulum (under the influence of friction):

   
   \theta(t) = \theta_0 e^{-\gamma t} \cos(\omega t)
   

   Where:

   • \gamma = damping coefficient

7. For a physical pendulum (a rigid body swinging about a pivot):

   
   T = 2\pi \sqrt{\frac{I}{mgd}}
   

   Where:

   • I = moment of inertia of the object
   • d = distance from the pivot point to the center of mass

These are the key formulas for general pendulum motion, covering simple and physical pendulums, damped motion, and energy analysis.

Dampened Springs

The damping coefficient, denoted by \gamma, quantifies the rate at which the amplitude of oscillation decreases over time due to resistive forces (like friction or air resistance) acting on the system. It depends on factors like the properties of the medium (air, fluid, etc.) and the characteristics of the object (shape, surface area, etc.).

The damping force is typically modeled as:

F_{\text{damping}} = -\gamma v

where:

• F_{\text{damping}} is the damping force,
• \gamma is the damping coefficient, and
• v is the velocity of the oscillating mass.

The damping coefficient influences how quickly the system loses energy. The larger the damping coefficient, the faster the oscillations decay.

Types of Damping:

1. Underdamped: If the damping is weak (\gamma < \omega_0), the system still oscillates, but with decreasing amplitude. The system's motion is described by:

   
   x(t) = A e^{-\gamma t} \cos(\omega' t + \phi)
   

2. Critically damped: If the damping is strong enough to prevent oscillations, but not too strong to cause the system to return to equilibrium too slowly, we have the critical damping condition: \gamma = \omega_0.

3. Overdamped: If the damping is too strong (\gamma > \omega_0), the system returns to equilibrium without oscillating, but more slowly than in the critically damped case.

Relation to Natural Frequency:

The damped angular frequency \omega' is related to the natural frequency \omega_0 by:

\omega' = \sqrt{\omega_0^2 - \gamma^2}

This shows that as \gamma increases, \omega' decreases, meaning the system oscillates less rapidly. If \gamma is large enough (in the overdamped case), there will be no oscillations at all.

In summary, the damping coefficient \gamma plays a key role in determining the rate at which a mass-spring system loses energy and the nature of its oscillations.

Snell's Law:

Snell's Law describes the relationship between the angles of incidence and refraction when a wave passes through the boundary between two media with different refractive indices. It is given by:

n_1 \sin(\theta_1) = n_2 \sin(\theta_2)

Where:

• n_1 and n_2 are the refractive indices of the two media,
• \theta_1 and \theta_2 are the angles of incidence and refraction, respectively, measured from the normal.

Snell's Law explains how light bends when it transitions from one medium to another (e.g., air to water). If the refractive index increases (e.g., from air to glass), the light slows down and bends towards the normal. Conversely, if the refractive index decreases (e.g., from water to air), light bends away from the normal.

Doppler Effect:

The Doppler Effect refers to the change in frequency (or wavelength) of a wave as observed by someone moving relative to the wave source. For sound waves, the observed frequency f' is given by:

f' = f \left( \frac{v \pm v_{\text{observer}}}{v \pm v_{\text{source}}} \right)

Where:

• f' is the observed frequency
• f is the emitted frequency,
• v is the wave speed in the medium,
• v_{\text{observer}} is the velocity of the observer relative to the medium,
• v_{\text{source}} is the velocity of the source relative to the medium.

The sign conventions depend on whether the observer and source are moving towards or away from each other:

• Moving towards each other: use the "+" sign in the numerator and the "-" in the denominator.
• Moving apart: use the "-" sign in the numerator and the "+" in the denominator.

In the case of light, the relativistic Doppler shift formula is used, where the frequency shift is influenced by both the motion and the speed of light.

The Doppler Effect explains phenomena such as the change in pitch of a passing car or the redshift/blueshift observed in the light from stars and galaxies. Sure! Here's a combined explanation of the path difference in the double-slit experiment, with the math expressed using LaTeX notation:

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Young's Experiment

In the double-slit experiment, light passes through two slits and creates an interference pattern on a screen. The path difference refers to the difference in the distances that the light waves travel from each of the two slits to a particular point on the screen.

What is Path Difference?

1. Two light waves are emitted from the two slits, and they travel towards a point on the screen.
2. The distance traveled by the light from the first slit to that point is different from the distance traveled by the light from the second slit.
3. The path difference is simply the difference between these two distances.

How Path Difference Affects the Interference Pattern:

• Constructive Interference: This occurs when the path difference is a whole number multiple of the wavelength, i.e.,

   
  \Delta y = m\lambda \quad \text{(where $m$ is an integer)} 
  

  In this case, the waves arrive in phase and interfere constructively, creating a bright fringe on the screen.

• Destructive Interference: This happens when the path difference is an odd multiple of half the wavelength, i.e.,

   
  \Delta y = \left( m + \frac{1}{2} \right) \lambda \quad \text{(where $m$ is an integer)} 
  

  Here, the waves arrive out of phase and interfere destructively, resulting in a dark fringe.

Formula for Path Difference:

In the double-slit experiment, the path difference for a point on the screen at an angle ( \theta ) can be expressed as:

\Delta y = d \sin \theta

Where:

• \Delta y = path difference
• d = distance between the two slits
• \theta = angle between the central maximum (the center of the pattern) and the point on the screen
• \lambda = wavelength of the light

At the central maximum (the middle bright fringe), the path difference is zero because the light from both slits travels the same distance. As you move away from the center, the path difference increases, affecting the type of interference (constructive or destructive) at each point on the screen.

How Path Difference Relates to Intensity:

The intensity of the light at a point on the screen in the double-slit experiment depends on the path difference. The intensity for constructive and destructive interference can be written as:

For constructive interference (bright fringes):

I_m = I_0 \cos^2\left(\frac{m\pi d}{\lambda D}\right)

For destructive interference (dark fringes):

I = I_0 \cos^2\left(\frac{\pi \Delta y}{\lambda}\right)

Where:

• I_m = intensity at the $ m $-th fringe
• I_0 = maximum intensity (at the central maximum)
• m = fringe order (1st, 2nd, etc.)
• \Delta y = path difference
• D = distance from the slits to the screen

Summary:

• The path difference is the difference in the distances traveled by the two light waves from the slits to a point on the screen.

• If the path difference is a whole multiple of the wavelength, you get constructive interference (bright fringe), and if it's an odd multiple of half the wavelength, you get destructive interference (dark fringe).

• The fringe width depends on the wavelength and the geometry of the setup and can be calculated using:

   
  \beta = \frac{\lambda D}{d}
  

  Where:

  • \beta = fringe width (distance between adjacent fringes)
  • \lambda = wavelength of the light
  • D = distance from the slits to the screen
  • d = distance between the two slits

Final equations

Time to combine everything together. We'll continue to use the variable definitions from above, but include y_m, or the distance between the center, m=0 and some bright fringe of order m. When calculating, it can also be assumed that (for small angles) \sin\theta\approx\theta.

\boxed{d\sin\theta=m\lambda}

\boxed{y_md=Dm\lambda}