- •Preface
- •Textbook Layout and Design
- •Preliminaries
- •See, Do, Teach
- •Other Conditions for Learning
- •Your Brain and Learning
- •The Method of Three Passes
- •Mathematics
- •Summary
- •Homework for Week 0
- •Summary
- •1.1: Introduction: A Bit of History and Philosophy
- •1.2: Dynamics
- •1.3: Coordinates
- •1.5: Forces
- •1.5.1: The Forces of Nature
- •1.5.2: Force Rules
- •Example 1.6.1: Spring and Mass in Static Force Equilibrium
- •1.7: Simple Motion in One Dimension
- •Example 1.7.1: A Mass Falling from Height H
- •Example 1.7.2: A Constant Force in One Dimension
- •1.7.1: Solving Problems with More Than One Object
- •Example 1.7.4: Braking for Bikes, or Just Breaking Bikes?
- •1.8: Motion in Two Dimensions
- •Example 1.8.1: Trajectory of a Cannonball
- •1.8.2: The Inclined Plane
- •Example 1.8.2: The Inclined Plane
- •1.9: Circular Motion
- •1.9.1: Tangential Velocity
- •1.9.2: Centripetal Acceleration
- •Example 1.9.1: Ball on a String
- •Example 1.9.2: Tether Ball/Conic Pendulum
- •1.9.3: Tangential Acceleration
- •Homework for Week 1
- •Summary
- •2.1: Friction
- •Example 2.1.1: Inclined Plane of Length L with Friction
- •Example 2.1.3: Find The Minimum No-Skid Braking Distance for a Car
- •Example 2.1.4: Car Rounding a Banked Curve with Friction
- •2.2: Drag Forces
- •2.2.1: Stokes, or Laminar Drag
- •2.2.2: Rayleigh, or Turbulent Drag
- •2.2.3: Terminal velocity
- •Example 2.2.1: Falling From a Plane and Surviving
- •2.2.4: Advanced: Solution to Equations of Motion for Turbulent Drag
- •Example 2.2.3: Dropping the Ram
- •2.3.1: Time
- •2.3.2: Space
- •2.4.1: Identifying Inertial Frames
- •Example 2.4.1: Weight in an Elevator
- •Example 2.4.2: Pendulum in a Boxcar
- •2.4.2: Advanced: General Relativity and Accelerating Frames
- •2.5: Just For Fun: Hurricanes
- •Homework for Week 2
- •Week 3: Work and Energy
- •Summary
- •3.1: Work and Kinetic Energy
- •3.1.1: Units of Work and Energy
- •3.1.2: Kinetic Energy
- •3.2: The Work-Kinetic Energy Theorem
- •3.2.1: Derivation I: Rectangle Approximation Summation
- •3.2.2: Derivation II: Calculus-y (Chain Rule) Derivation
- •Example 3.2.1: Pulling a Block
- •Example 3.2.2: Range of a Spring Gun
- •3.3: Conservative Forces: Potential Energy
- •3.3.1: Force from Potential Energy
- •3.3.2: Potential Energy Function for Near-Earth Gravity
- •3.3.3: Springs
- •3.4: Conservation of Mechanical Energy
- •3.4.1: Force, Potential Energy, and Total Mechanical Energy
- •Example 3.4.1: Falling Ball Reprise
- •Example 3.4.2: Block Sliding Down Frictionless Incline Reprise
- •Example 3.4.3: A Simple Pendulum
- •Example 3.4.4: Looping the Loop
- •3.5: Generalized Work-Mechanical Energy Theorem
- •Example 3.5.1: Block Sliding Down a Rough Incline
- •Example 3.5.2: A Spring and Rough Incline
- •3.5.1: Heat and Conservation of Energy
- •3.6: Power
- •Example 3.6.1: Rocket Power
- •3.7: Equilibrium
- •3.7.1: Energy Diagrams: Turning Points and Forbidden Regions
- •Homework for Week 3
- •Summary
- •4.1: Systems of Particles
- •Example 4.1.1: Center of Mass of a Few Discrete Particles
- •4.1.2: Coarse Graining: Continuous Mass Distributions
- •Example 4.1.2: Center of Mass of a Continuous Rod
- •Example 4.1.3: Center of mass of a circular wedge
- •4.2: Momentum
- •4.2.1: The Law of Conservation of Momentum
- •4.3: Impulse
- •Example 4.3.1: Average Force Driving a Golf Ball
- •Example 4.3.2: Force, Impulse and Momentum for Windshield and Bug
- •4.3.1: The Impulse Approximation
- •4.3.2: Impulse, Fluids, and Pressure
- •4.4: Center of Mass Reference Frame
- •4.5: Collisions
- •4.5.1: Momentum Conservation in the Impulse Approximation
- •4.5.2: Elastic Collisions
- •4.5.3: Fully Inelastic Collisions
- •4.5.4: Partially Inelastic Collisions
- •4.6: 1-D Elastic Collisions
- •4.6.1: The Relative Velocity Approach
- •4.6.2: 1D Elastic Collision in the Center of Mass Frame
- •4.7: Elastic Collisions in 2-3 Dimensions
- •4.8: Inelastic Collisions
- •Example 4.8.1: One-dimensional Fully Inelastic Collision (only)
- •Example 4.8.2: Ballistic Pendulum
- •Example 4.8.3: Partially Inelastic Collision
- •4.9: Kinetic Energy in the CM Frame
- •Homework for Week 4
- •Summary
- •5.1: Rotational Coordinates in One Dimension
- •5.2.1: The r-dependence of Torque
- •5.2.2: Summing the Moment of Inertia
- •5.3: The Moment of Inertia
- •Example 5.3.1: The Moment of Inertia of a Rod Pivoted at One End
- •5.3.1: Moment of Inertia of a General Rigid Body
- •Example 5.3.2: Moment of Inertia of a Ring
- •Example 5.3.3: Moment of Inertia of a Disk
- •5.3.2: Table of Useful Moments of Inertia
- •5.4: Torque as a Cross Product
- •Example 5.4.1: Rolling the Spool
- •5.5: Torque and the Center of Gravity
- •Example 5.5.1: The Angular Acceleration of a Hanging Rod
- •Example 5.6.1: A Disk Rolling Down an Incline
- •5.7: Rotational Work and Energy
- •5.7.1: Work Done on a Rigid Object
- •5.7.2: The Rolling Constraint and Work
- •Example 5.7.2: Unrolling Spool
- •Example 5.7.3: A Rolling Ball Loops-the-Loop
- •5.8: The Parallel Axis Theorem
- •Example 5.8.1: Moon Around Earth, Earth Around Sun
- •Example 5.8.2: Moment of Inertia of a Hoop Pivoted on One Side
- •5.9: Perpendicular Axis Theorem
- •Example 5.9.1: Moment of Inertia of Hoop for Planar Axis
- •Homework for Week 5
- •Summary
- •6.1: Vector Torque
- •6.2: Total Torque
- •6.2.1: The Law of Conservation of Angular Momentum
- •Example 6.3.1: Angular Momentum of a Point Mass Moving in a Circle
- •Example 6.3.2: Angular Momentum of a Rod Swinging in a Circle
- •Example 6.3.3: Angular Momentum of a Rotating Disk
- •Example 6.3.4: Angular Momentum of Rod Sweeping out Cone
- •6.4: Angular Momentum Conservation
- •Example 6.4.1: The Spinning Professor
- •6.4.1: Radial Forces and Angular Momentum Conservation
- •Example 6.4.2: Mass Orbits On a String
- •6.5: Collisions
- •Example 6.5.1: Fully Inelastic Collision of Ball of Putty with a Free Rod
- •Example 6.5.2: Fully Inelastic Collision of Ball of Putty with Pivoted Rod
- •6.5.1: More General Collisions
- •Example 6.6.1: Rotating Your Tires
- •6.7: Precession of a Top
- •Homework for Week 6
- •Week 7: Statics
- •Statics Summary
- •7.1: Conditions for Static Equilibrium
- •7.2: Static Equilibrium Problems
- •Example 7.2.1: Balancing a See-Saw
- •Example 7.2.2: Two Saw Horses
- •Example 7.2.3: Hanging a Tavern Sign
- •7.2.1: Equilibrium with a Vector Torque
- •Example 7.2.4: Building a Deck
- •7.3: Tipping
- •Example 7.3.1: Tipping Versus Slipping
- •Example 7.3.2: Tipping While Pushing
- •7.4: Force Couples
- •Example 7.4.1: Rolling the Cylinder Over a Step
- •Homework for Week 7
- •Week 8: Fluids
- •Fluids Summary
- •8.1: General Fluid Properties
- •8.1.1: Pressure
- •8.1.2: Density
- •8.1.3: Compressibility
- •8.1.5: Properties Summary
- •Static Fluids
- •8.1.8: Variation of Pressure in Incompressible Fluids
- •Example 8.1.1: Barometers
- •Example 8.1.2: Variation of Oceanic Pressure with Depth
- •8.1.9: Variation of Pressure in Compressible Fluids
- •Example 8.1.3: Variation of Atmospheric Pressure with Height
- •Example 8.2.1: A Hydraulic Lift
- •8.3: Fluid Displacement and Buoyancy
- •Example 8.3.1: Testing the Crown I
- •Example 8.3.2: Testing the Crown II
- •8.4: Fluid Flow
- •8.4.1: Conservation of Flow
- •Example 8.4.1: Emptying the Iced Tea
- •8.4.3: Fluid Viscosity and Resistance
- •8.4.4: A Brief Note on Turbulence
- •8.5: The Human Circulatory System
- •Example 8.5.1: Atherosclerotic Plaque Partially Occludes a Blood Vessel
- •Example 8.5.2: Aneurisms
- •Homework for Week 8
- •Week 9: Oscillations
- •Oscillation Summary
- •9.1: The Simple Harmonic Oscillator
- •9.1.1: The Archetypical Simple Harmonic Oscillator: A Mass on a Spring
- •9.1.2: The Simple Harmonic Oscillator Solution
- •9.1.3: Plotting the Solution: Relations Involving
- •9.1.4: The Energy of a Mass on a Spring
- •9.2: The Pendulum
- •9.2.1: The Physical Pendulum
- •9.3: Damped Oscillation
- •9.3.1: Properties of the Damped Oscillator
- •Example 9.3.1: Car Shock Absorbers
- •9.4: Damped, Driven Oscillation: Resonance
- •9.4.1: Harmonic Driving Forces
- •9.4.2: Solution to Damped, Driven, Simple Harmonic Oscillator
- •9.5: Elastic Properties of Materials
- •9.5.1: Simple Models for Molecular Bonds
- •9.5.2: The Force Constant
- •9.5.3: A Microscopic Picture of a Solid
- •9.5.4: Shear Forces and the Shear Modulus
- •9.5.5: Deformation and Fracture
- •9.6: Human Bone
- •Example 9.6.1: Scaling of Bones with Animal Size
- •Homework for Week 9
- •Week 10: The Wave Equation
- •Wave Summary
- •10.1: Waves
- •10.2: Waves on a String
- •10.3: Solutions to the Wave Equation
- •10.3.1: An Important Property of Waves: Superposition
- •10.3.2: Arbitrary Waveforms Propagating to the Left or Right
- •10.3.3: Harmonic Waveforms Propagating to the Left or Right
- •10.3.4: Stationary Waves
- •10.5: Energy
- •Homework for Week 10
- •Week 11: Sound
- •Sound Summary
- •11.1: Sound Waves in a Fluid
- •11.2: Sound Wave Solutions
- •11.3: Sound Wave Intensity
- •11.3.1: Sound Displacement and Intensity In Terms of Pressure
- •11.3.2: Sound Pressure and Decibels
- •11.4: Doppler Shift
- •11.4.1: Moving Source
- •11.4.2: Moving Receiver
- •11.4.3: Moving Source and Moving Receiver
- •11.5: Standing Waves in Pipes
- •11.5.1: Pipe Closed at Both Ends
- •11.5.2: Pipe Closed at One End
- •11.5.3: Pipe Open at Both Ends
- •11.6: Beats
- •11.7: Interference and Sound Waves
- •Homework for Week 11
- •Week 12: Gravity
- •Gravity Summary
- •12.1: Cosmological Models
- •12.2.1: Ellipses and Conic Sections
- •12.4: The Gravitational Field
- •12.4.1: Spheres, Shells, General Mass Distributions
- •12.5: Gravitational Potential Energy
- •12.6: Energy Diagrams and Orbits
- •12.7: Escape Velocity, Escape Energy
- •Example 12.7.1: How to Cause an Extinction Event
- •Homework for Week 12
Week 10: The Wave Equation |
439 |
10.3: Solutions to the Wave Equation
In one dimension there are at least three distinct solutions to the wave equation that we are interested in. Two of these solutions propagate along the string – energy is transported from one place to another by the wave. The third is a stationary solution, in the sense that the wave doesn’t propagate in one direction or the other (not in the sense that the string doesn’t move). But first:
10.3.1: An Important Property of Waves: Superposition
The wave equation is linear, and hence it is easy to show that if y1(x, t) solves the wave equation and y2(x, t) (independent of y1) also solves the wave equation, then:
y(x, t) = Ay1(x, t) + By2(x, t) |
(910) |
solves the wave equation for arbitrary (complex) A and B.
This property of waves is most powerful and sublime.
10.3.2: Arbitrary Waveforms Propagating to the Left or Right
The first solution we can discern by noting that the wave equation equates a second derivative in time to a second derivative in space. Suppose we write the solution as f (u) where u is an unknown function of x and t and substitute it into the di erential equation and use the chain rule:
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with a simple solution:
u = x ± vt
(911)
(912)
(913)
(914)
What this tells us is that any function
y(x, t) = f (x ± vt) |
(915) |
satisfies the wave equation. Any shape of wave created on the string and propagating to the right or left is a solution to the wave equation, although not all of these waves will vanish at the ends of a string.
10.3.3: Harmonic Waveforms Propagating to the Left or Right
An interesting special case of this solution is the case of harmonic waves propagating to the left or right. Harmonic waves are simply waves that oscillate with a given harmonic frequency. For example:
y(x, t) = y0 sin(x − vt) |
(916) |
is one such wave. y0 is called the amplitude of the harmonic wave. But what sorts of parameters describe the wave itself? Are there more than one harmonic waves?
This particular wave looks like a sinusoidal wave propagating to the right (positive x direction). But this is not a very convenient parameterization. To better describe a general harmonic wave, we need to introduce the following quantities:
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Week 10: The Wave Equation |
•The frequency f . This is the number of cycles per second that pass a point or that a point on the string moves up and down.
•The wavelength λ. This the distance one has to travel down the string to return to the same point in the wave cycle at any given instant in time.
To convert x (a distance) into an angle in radians, we need to multiply it by 2π radians per wavelength. We therefore define the wave number:
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and write our harmonic solution as: |
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(918) |
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(919) |
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(920) |
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where we have used the following train of algebra in the last step: |
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(921) |
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and where we see that we have two ways to write v: |
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As before, you should simply know every relation in this set of algebraic relations between λ, k, f, ω, v to save time on tests and quizzes. Of course there is also the harmonic wave travelling to the left as well:
y(x, t) = y0 sin(kx + ωt). |
(923) |
A final observation about these harmonic waves is that because arbitrary functions can be expanded in terms of harmonic functions (e.g. Fourier Series, Fourier Transforms) and because the wave equation is linear and its solutions are superposable, the two solution forms above are not really distinct. One can expand the “arbitrary” f (x −vt) in a sum of sin(kx −ωt)’s for special frequencies and wavelengths. In one dimension this doesn’t give you much, but in two or more dimensions this process helps one compute the dispersion of the wave caused by the wave “spreading out” in multiple dimensions and reducing its amplitude.
10.3.4: Stationary Waves
The third special case of solutions to the wave equation is that of standing waves. They are especially apropos to waves on a string fixed at one or both ends. There are two ways to find these solutions from the solutions above. A harmonic wave travelling to the right and hitting the end of the string (which is fixed), it has no choice but to reflect. This is because the energy in the string cannot just disappear, and if the end point is fixed no work can be done by it on the peg to which it is attached. The reflected wave has to have the same amplitude and frequency as the incoming wave. What does the sum of the incoming and reflected wave look like in this special case?
Suppose one adds two harmonic waves with equal amplitudes, the same wavelengths and frequencies, but that are travelling in opposite directions:
y(x, t) = |
y0 (sin(kx − ωt) + sin(kx + ωt)) |
(924) |
= |
2y0 sin(kx) cos(ωt) |
(925) |
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A sin(kx) cos(ωt) |
(926) |
Week 10: The Wave Equation |
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(where we give the standing wave the arbitrary amplitude A). Since all the solutions above are independent of the phase, a second useful way to write stationary waves is:
y(x, t) = A cos(kx) cos(ωt) |
(927) |
Which of these one uses depends on the details of the boundary conditions on the string.
In this solution a sinusoidal form oscillates harmonically up and down, but the solution has some very important new properties. For one, it is always zero when x = 0 for all possible lambda:
y(0, t) = 0 |
(928) |
For a given λ there are certain other x positions where the wave vanishes at all times. These positions are called nodes of the wave. We see that there are nodes for any L such that:
y(L, t) = A sin(kL) cos(ωt) = 0 |
(929) |
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which implies that: |
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(930) |
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for n = 1, 2, 3, ...
Only waves with these wavelengths and their associated frequencies can persist on a string of length L fixed at both ends so that
y(0, t) = y(L, t) = 0 |
(932) |
(such as a guitar string or harp string). Superpositions of these waves are what give guitar strings their particular tone.
It is also possible to stretch a string so that it is fixed at one end but so that the other end is free to move – to slide up and down without friction on a greased rod, for example. In this case, instead of having a node at the free end (where the wave itself vanishes), it is pretty easy to see that the slope of the wave at the end has to vanish. This is because if the slope were not zero, the terminating rod would be pulling up or down on the string, contradicting our requirement that the rod be frictionless and not able to pull the string up or down, only directly to the left or right due to tension.
The slope of a sine wave is zero only when the sine wave itself is a maximum or minimum, so that the wave on a string free at an end must have an antinode (maximum magnitude of its amplitude) at the free end. Using the same standing wave form we derived above, we see that:
kL = |
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= π/2, 3π/2, 5π/2 . . . |
(933) |
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for a string fixed at x = 0 and free at x = L, or:
λ = |
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(934) |
2n − 1 |
for n = 1, 2, 3, ...
There is a second way to obtain the standing wave solutions that particularly exhibits the relationship between waves and harmonic oscillators. One assumes that the solution y(x, t) can be written as the product of a fuction in x alone and a second function in t alone:
y(x, t) = X(x)T (t) |
(935) |
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Week 10: The Wave Equation |
If we substitute this into the di erential equation and divide by y(x, t) we get:
d2y |
= X(x) |
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(936) |
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(938) |
where the last line follows because the second line equations a function of t (only) to a function of x (only) so that both terms must equal a constant. This is then the two equations:
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From this we see that: |
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(941) |
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(942) |
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so that the most general stationary solution can be written: |
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y(x, t) = y0 cos(kx + δ) cos(ωt + φ) |
(943) |
10.4: Reflection of Waves
We argued above that waves have to reflect of of the ends of stretched strings because of energy conservation. This is true independent of whether the end is fixed or free – in neither case can the string do work on the wall or rod to which it is a xed. However, the behavior of the reflected wave is di erent in the two cases.
Suppose a wave pulse is incident on the fixed end of a string. One way to “discover” a wave solution that apparently conserves energy is to imagine that the string continues through the barrier. At the same time the pulse hits the barrier, an identical pulse hits the barrier from the other, “imaginary” side.
Since the two pulses are identical, energy will clearly be conserved. The one going from left to right will transmit its energy onto the imaginary string beyond at the same rate energy appears going from right to left from the imaginary string.
However, we still have two choices to consider. The wave from the imaginary string could be right side up the same as the incident wave or upside down. Energy is conserved either way!
If the right side up wave (left to right) encounters an upside down wave (right to left) they will always be opposite at the barrier, and when superposed they will cancel at the barrier. This corresponds to a fixed string. On the other hand, if a right side up wave encounters a right side up wave, they will add at the barrier with opposite slope. There will be a maximum at the barrier with zero slope – just what is needed for a free string.
From this we deduce the general rule that wave pulses invert when reflected from a fixed boundary (string fixed at one end) and reflect right side up from a free boundary.
When two strings of di erent weight (mass density) are connected, wave pulses on one string are both transmitted onto the other and are generally partially reflected from the boundary. Computing