Physics 20 - Oscillatory Motion & Mechanical Waves Notes

Mechanical Waves

  • Mechanical waves transmit energy through a medium via vibrations.
  • We'll explore the nature, properties, and energy transmission of waves.

8.1: Properties of Waves

  • Wave: A pattern traveling through a medium via vibration, an oscillation carrying energy.
  • In a perfectly elastic medium, waves may exhibit Simple Harmonic Motion (SHM).
Terminology:
  • Wave Front: An imaginary line connecting points reached by a wave simultaneously.
  • Incident Wave: A wave front moving from the origin towards a barrier.
  • Reflected Wave: A wave front moving away from a barrier.
  • Wave Train: A continuous series of crests and troughs.
  • Point Source: A single disturbance point generating a circular wave.
  • Ray: Shows the direction of wave front motion.
Waves and Rays
  • Waves move at right angles to the crest line, indicated by a velocity vector or ray.
  • Ripple Tank Example
  • Diverging Rays: Spread out from the origin, indicating energy dispersal over a growing area.
Wave Properties
  • Amplitude: Maximum displacement from equilibrium.
  • Wavelength: Distance between two identical points on a wave (crest to crest, trough to trough), where the wave begins to repeat its motion.

Waves and Boundaries: Reflection

  • When a wave meets a boundary, energy is transferred and reflected.
Example 1:
  • A wave front hits a barrier at a 30(circ30^(circ angle.
  • The angle of incidence equals the angle of reflection: θ<em>i=θ</em>r\theta<em>i = \theta</em>r.

8.4: The Doppler Effect

  • Occurs when a wave source (sound, light) is moving.
Doppler Equation:
  • f<em>d=(v</em>wv<em>w±v</em>s)fsf<em>d = (\frac{v</em>w}{v<em>w \pm v</em>s}) f_s
  • fdf_d = Doppler frequency (apparent frequency).
  • vwv_w = Speed of sound in air (approximately 340m/s340 m/s).
  • vsv_s = Speed of source.
  • fsf_s = True frequency of source.
  • Use minus (-) when the source moves toward the observer.
  • Use plus (+) when the source moves away from the observer.
Example 1:
  • A train travels at 30.0m/s30.0 m/s, its whistle emits sound at 224Hz224 Hz.
  • Calculate the frequency change as the train passes.

Bow Waves

  • Bow wave forms at the bow of a boat moving through water.
  • It defines the outer limits of a boat's wake.
  • For airplanes, the bow wave is a pressure wave traveling at the speed of sound.

The Sound Barrier

  • Crests of bow waves arrive simultaneously, creating a sonic boom.
  • Sonic boom: Sound from shock waves of an object exceeding the speed of sound; sounds like an explosion.
  • The sound barrier is the transition point from the speed of sound to supersonic speed.

Chapter 7: Oscillatory Motion Conditions

  • Oscillatory Motion: Motion with a constant period for each cycle.
Example 1:
  • Find the frequency of an engine with a piston oscillating at a period of 0.0625s0.0625 s.

  • T=tcyclesT = \frac{t}{cycles} (Period)

  • f=cyclessf = \frac{cycles}{s} (Frequency)

  • f=1Tf = \frac{1}{T}

  • f=10.0625=16.0Hzf = \frac{1}{0.0625} = 16.0 Hz

8.3: Superposition and Interference

  • Waves interact momentarily, then continue unaffected.
Principle of Superposition:
  • The displacement caused by multiple waves is the algebraic sum of individual amplitudes.
  • Constructive Interference
  • Destructive Interference

Standing Waves

  • Observed on strings fixed at both ends.
  • Resonant frequencies: Frequencies at which standing waves occur for a given string length.
  • Incident and reflected waves interfere.
  • Constructive interference (antinodes) where crests meet crests, troughs meet troughs.
  • Destructive interference (nodes) where crests meet troughs.
  • Nodes: Points with minimal displacement.
  • Antinodes: Points with maximum displacement.

Resonating Air Columns

  • Wind instruments use resonance for sound production.
Closed Columns:
  • Node at the closed end.
  • Antinode at the open end.
  • l=14λl = \frac{1}{4} \lambda
  • λ=4l\lambda = 4l
  • v=fλv = f \lambda
  • f=vλf = \frac{v}{\lambda}
  • Fundamental Frequency/1st Harmonic: f=v4lf = \frac{v}{4l}
  • Only odd harmonics produce sound in closed columns.
Open Columns:
  • Antinodes at both ends.
  • l=λ2l = \frac{\lambda}{2}
  • λ=2l\lambda = 2l
  • Fundamental Frequency/1st Harmonic: f=v2lf = \frac{v}{2l}
  • All harmonics produce sound in open columns.
  • vv is the speed of sound (340m/s340 m/s).
Example 1:
  • Open-ended organ column length l=3.6ml = 3.6 m.

  • Determine the wavelength of the fundamental frequency: λ=2l=2(3.6)=7.2m\lambda = 2l = 2(3.6) = 7.2 m.

  • Determine the frequency: f=vλ=3467.2=48Hzf = \frac{v}{\lambda} = \frac{346}{7.2} = 48 Hz.

  • The third harmonic: f=3(v2l)=3(3462(3.6))=144Hzf = 3(\frac{v}{2l}) = 3(\frac{346}{2(3.6)}) = 144 Hz

  • Increasing column length lowers the fundamental frequency.

Interference Patterns

8.2: Transverse and Longitudinal Waves

Transverse Waves:
  1. Classical wave shape.
  2. Particles move perpendicular to wave direction.
  3. Example: Ripples on a pond.
Longitudinal Waves:
  1. Displacement parallel to wave propagation.
  2. Series of compressions and rarefactions.
  3. Example: Sound waves.

Waves, Energy, and the Medium

  • Solids store elastic potential energy via longitudinal or transverse stretching.
  • Fluids store elastic potential energy via compression; waves are pressure waves.
  • Transverse waves in liquids are surface waves due to gravitational potential energy.

Pulse and the Universal Wave Equation

  • Pulse: A single crest or trough; length is half the wavelength.
  • Amplitude increases energy.
  • Speed from Universal Wave Equation: v=lΔtv = \frac{l}{\Delta t}. When l equals lambda, we get v=λTv = \frac{\lambda}{T}, therefore we derive v=λfv=\lambda f
  • λ\lambda = Wavelength (m)
  • TT = Period (s)
  • ff = Frequency (Hz)
Example 1:
  • Wrist flick creates a pulse in 0.54s0.54 s with a length of 0.78m0.78 m.
  • v=lt=0.780.54=1.4m/sv = \frac{l}{t} = \frac{0.78}{0.54} = 1.4 m/s.
Example 2:
  • Speed of light: 3.00×108m/s3.00 \times 10^8 m/s.
  • Wavelength of red light: 700nm700 nm.
  • f=vλ=3.00×108700×109=4.29×1014Hzf = \frac{v}{\lambda} = \frac{3.00 \times 10^8}{700 \times 10^{-9}} = 4.29 \times 10^{14} Hz.

Waves and Boundaries

  • Waves change velocity when moving between media.
  • Constant source frequency causes changes in velocity and wavelength: v=λfv = \lambda f.
Change in Medium Density:
  • Denser to less dense: Reflected the same way.
  • Less dense to denser: Reflects inverted.
Change in Depth:
  • Waves slow in shallow water, decreasing wavelength (constant frequency).
String Attached to…
  • Fixed end: Inverted reflection.
  • Loose end: Reflection is the same way.

7.2: Simple Harmonics

Hooke’s Law:
  • Deformation is proportional to force.
  • Spring deformation relates directly to applied force.
  • Spring stiffness (spring constant) is represented by kk.
  • Equation: F=kXF = kX
Hooke’s Law Example 1:
  • Plot force vs. displacement to find the spring constant.
  • k=ΔFΔx=2N/mk = \frac{\Delta F}{\Delta x} = 2 N/m

The Restoring Force

  • Spring exerts equal and opposite force.
  • Restoring force tries to return the spring to equilibrium.
  • Symbol: FrF_r.
  • F<em>a=F</em>rF<em>a = -F</em>r
Example 2:
  • k=30.0N/mk = 30.0 N/m, x=1.50mx = 1.50 m.
  • Fr=kx=(30)(1.5)=45.0NF_r = -kx = -(30)(-1.5) = 45.0 N

Simple Harmonic Motion (SHM) of Horizontal Mass-Spring System

  • SHM: Motion about equilibrium caused by restoring force proportional to displacement.
  • Horizontal mass-spring system oscillates with SHM.
Key Points:
  • Equilibrium position:
    • x=0x = 0, Fr=0F_r = 0, a=0a = 0, v=maxv = max
  • Max displacement (+A):
    • x=+max=+Ax = +max = +A, Fr=maxF_r = -max, a=maxa = -max, v=0v = 0
  • Max displacement (-A):
    • x=max=Ax = -max = -A, Fr=+maxF_r = +max, a=+maxa = +max, v=0v = 0

Simple Harmonic Motion of Vertical Mass-Spring System

  • Similar to horizontal, but with gravitational force (FgF_g).
Example 3:
  • Mass of 510g510 g stretches spring 0.500m0.500 m.

Simple Harmonic Motion of a Pendulum (small angles)

  • Exhibits SHM for small displacements.

Motion with Large Amplitudes

  • Non-linear graph; restoring force not proportional to displacement.
Example 4:
  • Find the restoring force for a pendulum bob of mass 100.0g100.0 g at an angle of 10.010.0^\circ.

7.3: Position, Velocity, Acceleration & Time Relationships

Acceleration of a Mass-Spring System
  • F<em>netF<em>{net} and F</em>sF</em>s are used to describe the force.
  • a=(km)xa = -(\frac{k}{m})x (Instantaneous acceleration)
  • Maximum acceleration occurs at maximum displacement (A).
  • Acceleration decreases as it approaches equilibrium.
  • Acceleration and displacement are in opposite directions.
Maximum Speed of a Mass-Spring System
  • Total mechanical energy is constant in an isolated system.
  • E<em>m=E</em>k+EpE<em>m = E</em>k + E_p
  • E<em>pmax=E</em>kmaxE<em>{pmax} = E</em>{kmax}
  • 12kA2=12mVm2\frac{1}{2} kA^2 = \frac{1}{2} mV_m^2
Example 1:
  • 100.0g100.0 g mass, k=1.014N/mk = 1.014 N/m, displacement of 15.0cm15.0 cm [up]: a=(km)xa = -(\frac{k}{m})x

Period of A Mass-Spring System

  • Circular motion (CM) describes Simple Harmonic Motion (SHM).
Three Conditions:
  1. Radius (r) of CM equals amplitude of SHM.
  2. Mass in CM with constant speed.
  3. Periods of CM and SHM are the same.
  • Equation for Period:
    • T=2πmkT = 2 \pi \sqrt{\frac{m}{k}}
  • Maximum speed of mass-spring system = speed of CM.
Example 2:
  • Amplitude is 12.25cm12.25 cm, maximum speed is 5.13m/s5.13 m/s, k=5.03N/mk = 5.03 N/m

The Period of A Pendulum (angles 15o or less)

  • Pendulum's mass does not affect the period.
The Pendulum and Gravitational Field Strength
  1. g varies with altitude and latitude.
  2. g=9.81N/kgg = 9.81 N/kg at Earth's average radius.
  3. Pendulums determine exact gg using the period formula.
The Pendulum's Period Equation Formula.
  • T=2πLgT = 2\pi\sqrt{\frac{L}{g}}
Example 3:
  • Find gravitational field strength at Mount Everest (altitude 8954.0m8954.0 m), pendulum length 1.00m1.00 m, period 2.01s2.01 s.