Oscillations and Waves Notes

6.1 Simple Harmonic Motion

Outline:
- Oscillations
- Example: Rocking chair
- Simple harmonic motion (SHM)
- SHM position
- Example: Position in SHM
- SHM velocity & acceleration
- Example: X-vs-t graph for SHM
Reference: Walker 13.1–13.3
Learning Objectives:
- Utilize frequency and amplitude to describe SHM. (Examples 1 & 2)
- Utilize formulae for SHM position as a function of time. (Example 2)
- Demonstrate conceptual understanding of the kinematics of SHM. (Example 3)

Oscillations

Periodic Motion: Any motion that repeats itself over time.
Period (T): Elapsed time for one cycle. SI unit: seconds (s)
Frequency (f): Number of cycles per unit time. SI unit: Hertz (Hz), where $f = s^{-1}$
Relationship between period and frequency: $f = \frac{1}{T}$
Oscillation: A simple form of periodic motion where an object moves back-and-forth between fixed values.
Examples: Heartbeat (periodic motion), pendulum of a grandfather clock (oscillates).

Example: Rocking Chair

A granny completes 12 rocks in 24 seconds.

Period (A): $T = \frac{24 \text{ s}}{12 \text{ rocks}} = 2 \text{ s}$
Frequency (B): $f = \frac{1}{T} = \frac{1}{2 \text{ s}} = 0.5 \text{ Hz}$

Simple Harmonic Motion (SHM)

The simplest type of oscillation.
Key ingredients:
- Restoring Force: Always drives the system toward equilibrium.
- Inertia: Causes the system to "overcorrect."
SHM is a universal behavior (examples: spring-mass, pendulum, buoy in water, trampolines, vibrating molecules, stellar surfaces, man-made structures).
*Example: Spring-mass system on a smooth surface with the spring force as the restoring force and the mass providing inertia.

SHM Position

The position of an object in SHM is sinusoidal, described by two parameters:
- Amplitude (A): Maximum displacement from equilibrium (x = 0).
- Period (T): Elapsed time for one cycle.
Position as a function of time:
$x(t) = A \cos\left(\frac{2\pi}{T} t\right)$
*Calculator must be in radian mode.

Example: Position in SHM

An air-track cart attached to a spring completes one oscillation every 2.4 s. At t = 0, the cart is released from rest at a distance of 0.10 m from its equilibrium position.

A. Write the position of the cart as a function of time.
- Given: $T = 2.4 \text{ s}$ , $A = 0.10 \text{ m}$
- $x(t) = (0.10 \text{ m}) \cos\left(\frac{2\pi}{2.4 \text{ s}} t\right)$
B. What is the position of the cart at t = 0.3 s?
- $x(0.3 \text{ s}) = (0.10 \text{ m}) \cos\left(\frac{2\pi}{2.4 \text{ s}} (0.3 \text{ s})\right) \approx 0.0707 \text{ m}$
C. What is the first time the cart is at the position x = 0 m?
- Solve $0 = (0.10 \text{ m}) \cos\left(\frac{2\pi}{2.4 \text{ s}} t\right)$ for $t$
- $\frac{2\pi}{2.4 \text{ s}} t = \frac{\pi}{2}$
- $t = \frac{2.4 \text{ s}}{4} = 0.6 \text{ s}$
D. Determine the position of the cart at t = 2.7 s without using a calculator.
- $t = 2.7 \text{ s} = 2.4 \text{ s} + 0.3 \text{ s}$ , so it's $0.3 \text{ s}$ into the next cycle, which is the same as $t = 0.3 \text{ s}$ in the first cycle.
- $x(2.7 \text{ s}) = x(0.3 \text{ s}) \approx 0.0707 \text{ m}$

SHM Velocity & Acceleration

Velocity and acceleration in SHM are also sinusoidal.
For SHM with $x(t) = A \cos\left(\frac{2\pi}{T} t\right)$ :
- Velocity: $v_x(t) = - \frac{2\pi}{T} A \sin\left(\frac{2\pi}{T} t\right)$
- Acceleration: $a_x(t) = - \left(\frac{2\pi}{T}\right)^2 A \cos\left(\frac{2\pi}{T} t\right)$

Example: X-vs-T Graph for SHM

A. Position:
- Maximum: Points A, E
- Minimum: Point C
- Zero: Points B, D, F
B. Velocity:
- Maximum: Point F
- Minimum: Point B
- Zero: Points A, C, E
C. Acceleration:
- Maximum: Point C
- Minimum: Points A, E
- Zero: Points B, D, F

6.2 Applications of SHM

Outline:
- The spring-mass system
- Example: Spring-mass system
- The simple pendulum
- Example: Grandfather clock
- Concept question
- The physical pendulum
- Example: Hula hoop
Reference: Walker 13.4, 13.6
Learning Objectives:
- For the spring-mass, simple pendulum, and physical pendulum systems:
  - Identify the restoring force & inertia of the system.
  - Qualitatively predict how changes to system parameters affect the period of oscillation (Example 1, Concept question).
  - Apply the formulae for period of oscillation to solve problems (Examples 2 & 3).

The Spring-Mass System

The period of oscillation depends on the "eternal war" between restoring force & inertia.
increase $m$ (inertia) ⇒ increase $T$
Increase $k$ (restoring force) ⇒ decrease $T$
Period does not depend on amplitude ( $A$ )!
Formula for the period:
$T = 2\pi \sqrt{\frac{m}{k}}$

Example: Spring-Mass System

a) Increase force constant (k) by a factor of 4: The period decreases.
b) Increase mass (m) by a factor of 4: The period increases.
c) Increase force constant (k) and mass (m) by a factor of 4: The period remains the same.
d) Increase amplitude (A) by a factor of 2: The period remains the same.

The Simple Pendulum

A simple pendulum is a model consisting of a point particle "bob" and a massless string or rod that swings back and forth in SHM.
Inertia: bob's mass ( $m$ or $I = mL^2$ )
Restoring force: tangential component of weight force on bob
For small angles ( $\theta$ ):
Period does not depend on amplitude or bob mass!
$T = 2\pi \sqrt{\frac{L}{g}}$

Example: Grandfather Clock

The pendulum in a grandfather clock is designed to take 1.00 s to swing in each direction, for a complete cycle.

Given: $T = 2.00 \text{ s}$
Solve for $L$ in $T = 2\pi \sqrt{\frac{L}{g}}$
$L = g \left(\frac{T}{2\pi}\right)^2 = (9.8 \text{ m/s}^2) \left(\frac{2.00 \text{ s}}{2\pi}\right)^2 \approx 0.993 \text{ m}$

The Physical Pendulum

A more realistic model of an object that swings back and forth at a small angle.
*Inertia: MOI of rigid body
*Restoring force: tangential component of weight force on body
*For small $\theta$ :
$T = 2\pi \sqrt{\frac{I}{mgL}}$
Where $I$ is the moment of inertia.
A physical pendulum is composed of a rigid body which swings about a pivot.

Example: Hula Hoop

A hula hoop with mass $m = 0.68$ kg and radius $R = 0.51$ m hangs from a peg in a garage.

*Moment of inertia of a hoop about a point on its rim is $I = 2mR^2$
$T = 2 \pi \sqrt{\frac{2mR^2}{mgR}} = 2 \pi \sqrt{\frac{2R}{g}} = 2 \pi \sqrt{\frac{2 (0.51 \text{ m})}{9.8 \text{ m/s}^2}} = 2.0 \text{ s}$

6.3 Transverse Waves

Outline:
- Waves
- Types of waves
- Wave properties
- Example: Wave graphs
- More wave properties
- Example: Brother & sister
- Concept question
Reference: Walker 14.1, 14.2
Learning Objectives:
- Distinguish between transverse and longitudinal waves.
- Identify and relate the properties of a wave, including amplitude, wavelength, period/frequency, and speed. (Example 1)
- Relate medium properties to wave speed. (Concept question)
- Utilize the formula for speed of a wave on a string. (Example 2)

Waves

A disturbance that propagates from one place to another.
Waves travel through a medium.
Waves are generated by oscillations.
Waves carry energy.
Ubiquitous in nature.
*Mechanical waves (described by Newton's Laws) are the focus.

Types of Waves

The motion of the wave pattern is different than the motion of the "pieces" of the medium.
- Transverse waves: Displacement of medium is perpendicular to the direction of wave motion.
- Longitudinal waves: Displacement of medium is parallel to the direction of wave motion.
- Often both types of waves occur simultaneously (e.g., water waves).

Wave Properties

Amplitude (A): Maximum distance the medium is displaced.
Period (T): Elapsed time of one cycle.
Wavelength ( $\lambda$ ): Length of one cycle along the wave. SI unit: meters (m).
Wave speed (v): Speed that wave pattern moves through the medium.
$v = \frac{\lambda}{T} = \lambda f$

Example: Wave Graphs

A. Amplitude: $A = 2 \text{ cm}$
B. Wavelength: $\lambda = 4 \text{ m}$
C. Period: $T = 2 \text{ s}$
D. Speed: $v = \frac{\lambda}{T} = \frac{4 \text{ m}}{2 \text{ s}} = 2 \text{ m/s}$
E. Direction of motion: $\pm \hat{x}$

Wave Speed

Wave amplitude & frequency are determined by source.
Wave speed is determined by properties of the medium.
Increase restoring force ⇒ increase $v$
Increase inertia ⇒ decrease $v$
$v = \lambda f$
You can choose a wave’s frequency or wavelength, but not both!
$v = \sqrt{\frac{F}{\mu}}$
*F: Tension
$\mu = \frac{m}{L}$
*Linear mass density
Waves on a string

Example: Brother & Sister

A brother and sister try to communicate with a string tied between two tin cans. The string is $7.6 \text{ m}$ long, has a mass of $29 \text{ g}$ , and is pulled taut with a tension of $18 \text{ N}$ .
$\mu = \frac{m}{L} = \frac{0.029 \text{ kg}}{7.6 \text{ m}} \approx 0.0038 \text{ kg/m}$
$v = \sqrt{\frac{F}{\mu}} = \sqrt{\frac{18 \text{ N}}{0.0038 \text{ kg/m}}} \approx 69 \text{ m/s}$
$t = \frac{L}{v} = \frac{7.6 \text{ m}}{69 \text{ m/s}} \approx 0.11 \text{ s}$

6.4 Sound & Doppler Effect

Outline:
- Sound waves
- Sound speed
- Frequency & pitch
- Concept question
- Doppler effect
- Example: Whistle stop
- Example: Angry birds
Reference: Walker 14.4, 14.6
Learning Objectives:
- Recognize sound waves as longitudinal waves.
- Be familiar with properties of sound waves, including sound speed, frequency, and pitch (Concept question).
- Demonstrate conceptual understanding of the Doppler effect (Example 1 & 2).
- Apply the general formula for the Doppler effect to solve problems (Example 1 & 2).

Sound Waves

Sound waves are longitudinal waves composed of traveling regions of compression and rarefaction.
Pieces of medium move in SHM.
Plots of various quantities (like density or pressure) show the same sinusoidal wave pattern of transverse waves.

Sound Speed

Wave speed is determined by properties of the medium.
Speed of sound in room temperature air: $v \approx 343 \text{ m/s}$
$v \sim \sqrt{\frac{\text{restoring force}}{\text{inertia}}}$
*Roughly 770 mi/hr.

Frequency & Pitch

Frequency of a sound wave determines the pitch of a sound.
Pitch ∝ frequency, so low pitch = small $f$ , high pitch = large $f$
Range of human hearing: 20 Hz - 20,000 Hz
Sound speed is constant in a medium.

Doppler Effect

When a source & observer have relative velocity, the observer will measure a different frequency than the source emits. These effects are known as the Doppler effect.

If observer & source move closer, then fo > fs
If observer & source move apart, then fo < fs
$fo = fs \left(\frac{v \pm vo}{v \mp vs}\right)$
Where:
$v$ is the speed of sound
* $vo$ is the speed of the observer * $vs$ is the speed of the source

Example: Whistle Stop

A train sounds its whistle as it approaches a tunnel in a cliff. The whistle produces a tone of $650.0 \text{ Hz}$ , and the train travels with a speed of $21.2 \text{ m/s}$ .

A. Find the frequency heard by an observer standing near the tunnel entrance.
The observer is stationary, so $vo = 0$ The source (train) is moving towards the observer, use the minus '-' sign in the denominator, $vs = 21.2 \text{ m/s}$
$f_o = (650.0 \text{ Hz}) \left(\frac{343 \text{ m/s}}{343 \text{ m/s} - 21.2 \text{ m/s}}\right) \approx 693 \text{ Hz}$
B. What frequency does the engineer hear?
Observer (engineer) is moving towards the source (reflected wave), use '+' sign in the numerator.
The frequency of the reflected wave equals the frequency heard by the observer at the cliff $693 \text{ Hz}$
$f_o = (693 \text{ Hz}) \left(\frac{343 \text{ m/s} + 21.2 \text{ m/s}}{343 \text{ m/s}}\right) \approx 736 \text{ Hz}$

Example: Angry Birds

Two fast and angry birds approach each other head-on. Alice has a speed of $6.00 \text{ m/s}$ and Bob has a speed of $3.00 \text{ m/s}$ . Alice sends out a 1350-Hz call toward Bob.

A. What is the frequency of sound detected by Bob?

$fo = fs \left(\frac{v + vo}{v - vs}\right) = 1350 \text{ Hz} \left(\frac{343 \text{ m/s} + 3.00 \text{ m/s}}{343 \text{ m/s} - 6.00 \text{ m/s}}\right) = 1350 \text{ Hz} \left(\frac{346}{337}\right) = 1386 \text{ Hz}$

B. What frequency of sound does Alice detect for this reflected wave?

$fo = fs \left(\frac{v + vo}{v - vs}\right) = 1386 \text{ Hz} \left(\frac{343 \text{ m/s} + 6.00 \text{ m/s}}{343 \text{ m/s} - 3.00 \text{ m/s}}\right) = 1386 \text{ Hz} \left(\frac{349}{340}\right) = 1423 \text{ Hz}$

6.5 Superposition & Interference

Outline:
- The Principle of Superposition
- Example: Wave collisions
- Interference
- Interference & sound waves
- Example: Dueling pianos
- Interference patterns
- Concept question
Reference: Walker 14.7
Learning Objectives:
- Apply the principle of superposition to determine the “collision” of waves on a string (Example 1).
- Identify constructive & destructive interference.
- Determine whether sound waves emitted by two sources constructively or destructively interfere (Example 2, concept question).

The Principle of Superposition

Superposition: when two or more waves are present simultaneously, the resultant disturbance is the algebraic sum of the individual waves.

Waves do not "interact" in usual sense — they pass through each other like ghosts! This has interesting consequences.
Principle of superposition applies in general for waves with small amplitude.

*Example of listening to an orchestra.

$y = y1 + y2$

Example: Wave Collisions

Pulses A, B, C, and D all travel at 10 m/s on the same string but in opposite directions.

Displacement of point P at t = 0.10 s

Only pulses C and D will be at point P at t = 0.10 s. Pulse C has a positive amplitude of 4.0 cm, while pulse D has a negative amplitude of -4.0 cm. The displacement will be y = 4.0 cm - 4.0 cm = 0 cm

Displacement of point P at t = 0.20 s

At t = 0.20s, pulses B and C will be at point P. The displacement will be y = -2.0 cm + 4.0 cm = 2.0 cm

Interference

Constructive interference: when waves combine to form a wave with amplitude equal to the sum of the individual wave amplitudes.
Destructive interference: when waves combine to form a wave with zero amplitude.
When waves combine they form interesting patterns called interference.

Interference & Sound Waves

Sound waves provide concrete examples of interference phenomena.
"In phase" sources: emit wave crests at the same time.

Constructive: $\Delta L = 0, \lambda, 2\lambda, \dots$
Destructive: $\Delta L = \frac{\lambda}{2}, \frac{3\lambda}{2}, \frac{5\lambda}{2}, \dots$

"Opposite phase" sources: one emits a wave when the other emits a trough.

Constructive: $\Delta L = \frac{\lambda}{2}, \frac{3\lambda}{2}, \frac{5\lambda}{2}, \dots$
Destructive: $\Delta L = 0, \lambda, 2\lambda, \dots$

Where $\Delta L = L2 - L1$

Example: Dueling Pianos

Two pianists perform at pianos positioned one directly behind the other with the audience directly in front of them. The pianists play a concert C $(523 \text{ Hz})$ in phase with one another. What is the smallest separation between the pianists that will produce constructive interference for the listener?

$v = 343 \text{ m/s}$ and $f = 523 \text{ Hz}$ , the wavelength is:

$\lambda = \frac{v}{f} \ \lambda = \frac{343 \text{ m/s}}{523 \text{ Hz}} = 0.656 \text{ m}$

So, the answer is $0.656 \text{ m}$ .

Interference Patterns

Constructive interference along this line, where crest meets crest.
Destructive interference along this line, where crest meets trough.

6.6 Standing Waves

Outline:
- Standing waves
- Transverse standing waves
- Example: Transverse standing waves
- Standing waves as interference
- Longitudinal standing waves
- Example: Longitudinal standing waves
Reference: Walker 14.8
Learning Objectives:
*Recognize standing waves as the result of wave interference.
Solve problems involving transverse standing waves (Example 1) & open- open longitudinal standing waves (Example 2).

Standing Waves

Standing waves are waves that oscillate (or “breath”) in time, but appear to have fixed position.
Standing waves are “trapped” in a region. Their shape depends upon the “boundary conditions” imposed at the ends.
Standing waves may be regarded interference between left-moving and right-moving waves.
Nodes: points that do not move
Anti-nodes: points of max displacement

Transverse Standing Waves

Harmonics: standing waves must “fit” inside the string — they must have nodes at the end points — so only certain wavelengths are allowed.
Harmonic number: # anti-nodes
Fundamental harmonic ( $n = 1$ ):
$\lambda1 = 2L, f1 = v/ {\lambda1} = v / 2L$ Higher harmonics: $n = 1,2,3,\dots$ $\lambdan = 2L/n, , fn = v/{\lambdan} = vn / 2L = n f_1$

*Where L is the length.

Example: Transverse Standing Waves

A 6.00-m long string sustains the standing wave pattern shown when the string is driven at a frequency of 60 Hz. What is the lowest possible frequency for standing waves on this string?

The third harmonic and number are given as well. This means that
v = 2Lf/n = (2)(6.0 m)(60 Hz) / 3 = 240 m/s.

Using that the first harmonic equals

f_1 = v/2L = (240 m/s) / (2 * 6.0 m) = 20.0 Hz.

Application: String Instruments

When a string is plucked, standing waves are generated, exciting mostly the fundamental harmonic.
Different string lengths have different fund. frequencies and thus different pitches.
$fn = v / {\lambdan} = nv / 2L = nf_1$

Standing Waves as Interference

Standing waves may be regarded as interference between left-moving and right-moving waves.
Right-moving wave + Left-moving wave = Standing wave.
Longitudinal Standing Waves

Longitudinal standing waves can form inside a cavity.
Examples: organ pipes, wind instruments, vocal chords, …
A cavity can have a open or closed boundaries.
- At open boundaries, standing waves have anti-nodes.
- At closed boundaries, standing waves have anti-nodes.

*Nodes: points that do not move.
*Anti-nodes: points of maximum displacement.

Longitudinal Standing Waves

Longitudinal standing waves with “open-open” boundaries have identical features as transverse waves with “closed-closed” boundaries.
Harmonic number: # nodes
$fn = v/{\lambdan} = nv / 2L = nf_1$

Example: Longitudinal Standing Waves

The frequency of the standing wave shown is 330 Hz. What is the length of the pipe?

The length of the pipe is given as

$L = 3*{\lambda}/4$ , and the frequency = 330 Hz (Given)

Frequency = v/{\lambda} = 330 (Given)

Therefore,

$L= 3v/4*330$ and

In air, v = 343 m/s

Hence,

$L = 3343/4330 = 0.78 m$