waves harmonic motion periodic motion repeat in a regular cycle –examples period t –time...

WavesWaves

Harmonic Motion

Periodic Motion

• Repeat in a regular cycle– Examples

• Period T– Time required to complete one cycle

• Frequency f – Cycles that occur in one second– Frequency is in hertz, Hz, which is 1/sec

Periodic Motion

• Springs as periodic motion• Always tries to return to equilibrium

• Hooke’s Law• F=-kx

• Potential Energy in a Spring• PEsp=1/2 kx^2

Periodic Motion

• Pendulum as simple harmonic motion• Always tries to return to equilibrium

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=11

Distance from equilibrium as a function of time

Wave PropertiesWave PropertiesTypes of Waves and Measuring Waves

Wave Properties• Amplitude

Maximum distance from equilibrium

Wave Properties• Peak-to-Peak Amplitude

Distance from Crest to Trough (or from low peak to high peak)

Wave Properties• Wavelength λ

Wave Properties• Period/frequency

– Essentially the same variable, just inverted• T = 1/f Seconds per cycle• f = 1/T Cycles per second

– (just remember frequency always has 1 sec as the denominator)

Cycles per second and seconds per cycle

Crash

Wave Properties• Phase

– A way to describe the relationship between two points on waves

¼ of the distance from the other wave

¼ x 360 = 90 degrees out of phase

Wave Properties

• Equations• Frequency: f = 1/T• Wavelength: λ = v/f • flambda: v = f λ (speed or velocity of a wave)

met

ers

x(meters)t(seconds)

¼ wavelength

Wave Properties

Problems

• In Class– Pg 397

• 64, 65, 66, 67, 68, 69, 71, 73, 75, 76, 77, 79, 80

• Homework– Pg 398

• 81, 82, 88, 89, 90, 97, 102

Wave BehaviorWave BehaviorMultiple waves, multiple boundaries

Wave Behavior

• Multiple waves can exist in the same place at the same time.

• Multiple particles cannot exist in the same place at the same time

• Note: This is where the dilemma occurs when waves and particles are not clearly separated.

Boundaries

• Incident waves– Waves that strike the boundary

• Reflected waves– Waves that return from the boundary

Boundary

Reflected Waves

• Fixed/Rigid– Inverted reflection

• Loose– Non-inverted reflection

• No end– No reflection

• Reflected waves do not change speed (ever). Speed is a function of the medium.

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=11

Principle of superposition

• The displacement of the medium caused by two or more waves is the algebraic sum of the displacements caused by the individual waves.

• Interference: The result of the superposition of two or more waves.

Interference

• Constructive– Result is bigger

• Destructive– Result is smaller

Interference

• Node: Has zero displacement, does not move at all.

• Antinode: Has maximum displacement.

What is the pattern? (# of nodes vs. antinodes)

Standing Wave

• Standing wave: Appears to be standing still. Is the interference of two or more waves.

Waves in two dimensions

Waves in two dimensions

Waves in two dimensions

• Wave front: A line that represents the crest or peak of a wave in 2D. Can be any shape (e.g. straight, circular, etc.).

Law of reflection

• Law of reflection– The angle of incidence is equal to the angle of

reflection

Refraction

• Refraction: The change in the direction of waves at the boundary between different media.– Examples are echoes and rainbows

Rules to Remember• Speed of a wave only changes when the medium

changes (i.e. water depth changes or wave crosses a boundary into a different medium).

• Speed will not change because of wavelength or frequency

• Frequency only changes when the source changes• Frequency will never change when the wave crosses a

boundary

v f

v flambda

Read the Graph

• Maximum Speed, Zero Force, No Acceleration Equilibrium Point

• Maximum Displacement, Maximum Amplitude, Maximum Force, Zero Velocity, No Motion

Things you need to know

• E=PE+KE• Work is Energy - work

done is equivalent to the change in potential energy

• All lengths need to be in meters

Problems

• Consider a 10.0-kg pendulum clock that has a period of 1s on Earth. If the clock is moved to a location where it weighs 98 N, how many minutes will the minute hand move in 1 h?– Write the equation for Earth– Write the equation for the new location– Set up the ratios

Problems

• Spring A with a spring constant of 253 N/m is stretched by a distance of 18.0 cm when a block is suspended from its end. An object is suspended from another spring B with a spring constant of 169 N/m. If the elastic potential energy in both the springs is the same, how far does spring B stretch?– Two parts– Find k with F=-kx– Compare with PE= ½ kx^2

Problems

• A 150-g object subject to a restoring force F = –kx is undergoing simple harmonic motion. Shown below is a plot of the potential energy, PE, of the object as a function of distance, x, from its equilibrium position. The object has a total mechanical energy of 0.3 J.a. What is the farthest the object moves along

the x-axis in the positive direction? Explain your reasoning.b. What is the object’s potential energy when its displacement is +4.0 cm from its quilibrium position?c. Determine the object’s kinetic energy when its position is x = –8.0 cm.d. What is the object’s speed at x = 0.0 cm?

Answer• a. Answer: 10.0 cm or 0.100 m. The total mechanical energy is E =

0.3 J, and total mechanical energy is given by E = KE + PE. Since the maximum potential energy cannot be greater than the total mechanical energy, the maximum potential energy is also 0.3 J. From the graph, PE = 0.3 J at x = 10.0 cm. Hence the maximum possible position in the +x-direction is 10.0 cm or 0.100 m. The particle stops at this point (i.e. the kinetic energy KE = 0 at this point) because all of the energy is in the form of potential energy.

• b. PE ≈ 0.05 J– Method 1:– Read off the value from the graph: PE ≈ 0.05 J– Method 2:– Since the restoring force is given by Hooke’s law F = –kx, we can use

the following equation for the potential energy of simple harmonic motion:

– PEsp = ½ kx^2– The maximum value of potential energy is PEmax = 0.3 J. This occurs

at a distance of xmax = 10.0 cm = 0.100 m. Use these values to determine the value of the spring constant k:

– k = 2PEmax/(x max^2) = 2(0.3J)/(0.100m)2 = 60M/n– Then, to determine the value of PE at x = 14.0 cm = 0.040 m:– PEsp = ½ kx^2 = ½ (60N/m)(0.040m)^2 = 0.05J

Answer• c. 0.1 J

– Method 1:– Read off the value from the graph: PE ≈ 0.2 J The total mechanical energy is

the sum of kinetic energy and potential energy:– E = KE + PE– Find the kinetic energy:– KE = E – PE = 0.3 J – 0.2 J = 0.1 J– Method 2:– Use the value of the spring constant k found in part b:– k = 60 N/m– Determine the value of PE at x = 28.0 cm = 20.080 m:– PEsp = ½ kx^2 = (60N/m)(20.080m)^2 = 0.2J– Then, find the kinetic energy:– E = KE + PE– KE = E – PE– = 0.3 J – 0.2 J = 0.1 J

• d. 2 m/s– At x = 0.0cm,PEsp = ½ kx^2 = 0– E = KE + PE = KE + 0– so– KE = E; ½ mv2 = E– Solving for the speed,– v = sqrt(2E/m)– v =sqrt[2(0.3J)/(0.15 kg)] = 2m/s

Problem

• Read the graph– Given: the incident wave travels at 1.0m/s– How much time has passed in graph b?– How far has the transmitted wave traveled in that

time?– What is the velocity of the transmitted wave in

graph b?

Problem• Given: 3 nodes and 2 antinodes in a 2-meter

span • What happens when the frequency is

doubled?– What is the wavelength?– Does the velocity change if the frequency changes?– Write an equation relating the first frequency and

wavelength to the second frequency and wavelength.

1 1 2 2v f f

Wave table or Ripple tank

Problems• In Class

– Pg 391• 27, 28, 29, 30,

• Homework– Pg 398

• 84, 85, 86, 87, 91, 93

Boundaries• Fixed• Loose• No end• Incident waves• Reflected waves: Do waves slow down after

reflection?• Principle of superposition• Wave interference• Node• Antinode• Standing waves• 2-D reflection• 2-D refraction• Normal• Law of reflection: angle of incidence is equal to angle

of reflection

wave-on-a-string_en.jar

http://phet.colorado.edu/simulations/sims.php?sim=Wave_on_a_String

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