Chapter 6 Flashcards
Amplitude
Max displacement from the undisturbed postion
Wavelength
Distance between two adjacent points that are in phase
Time period
Time to complete one wave cycle
Frequency
Number of cycles or oscillations per second. Number of wave crests passing a point per second
frequency =
1/Time period
Wave speed =
frequency * wavelength
Speed of sound (air)
330 m/s
Speed of light (vacuum)
3.0*10^8 m/s
Phasors
Rotating arrow, representing a wave. If wave left to right, rotates anti clockwise. Length is amplitude; time for one cycle is time period; rotates with frequency of wave. Projection of arrow onto vertical axis gives displacement of wave at that moment.
General wave equation
vertical displacement, y =
Asin(2pift) (A is amplitude; 2pi is radians; f is frequency; t is time passed)
or Asin(w*t); w is angular velocity
Angular velocity, w (omega) =
2pif
Superposition
Where 2 or more waves pass through one another, the total displacement at a point is equal to the vector sum of the individual displacements at that point
In phase
Phasors at same angle
Anti phase
Phasors at 180(deg) or 2*pi(rad)
Constructive interferance
Occurs when 2 waves are in phase; phasors at same angle; maximum is heard; path difference is n(lamda)
Destructive interferance
Occurs when 2 waves are in anti phase; phasors at 180(deg)/2pi(rad) ; minimum is observed; path difference is (2n+10.5(lamda)
To observe a stable interference pattern
Waves must be coherent
Coherent
Waves have a constant phase difference
Phase change on reflection
Transverse: phase change of 180(deg)/pi(rad) at a ‘rare-to-dense’ boundary (e.g. air to water/fixed end). No phase change from dense-to-rare boundary
Longitudinal: phase change of 180(deg)/pi(rad) at a dense-to-rare boundary (e.g. water to air/loose end). No phase change from rare-to-dense boundary
Standing wave
Superposition pattern of 2 identical waves, travelling in opposite directions
Identical waves
Same frequency, wavelength and speed; ideally same amplitude
Distance between adjacent nodes (or antinodes) on a standing wave
1/2(lamda)
Where are standing waves most clear
Close to the reflector, as the waves have traveled similar distances, and have lost similar amounts of energy, meaning their amplitude’s are of similar magnitude
Lloyd’s mirror
An example of superposition, and a way of measuring wavelength. Wave transmitter + receiver in front of a mirror. As mirror is moved from 1 maximum to the next maximum, path difference has changed by 1 wavelength
Huygen’s theory
Every point on a wave is a secondary source, so multiple wave lengths are emitted. The next wavefront is their superposition pattern
Fixed end standing waves, fundamental frequency
2 nodes, each at fixed end. Length, L is half the wavelength. frequency = f0
Fixed end standing waves, 3 nodes
2 nodes at each end, one in middle. Length is the wavelength. frequency = 2*f0
Fixed end standing waves, 4 nodes
2 nodes at each end, 2 in middle. Length is 1.5 wavelengths. frequency = 3*f0
Fixed end standing waves, 5 nodes
2 nodes at each end, 3 in middle. Length is 2 wavelengths. Frequency = 4*f0
One closed end standing wave
1 node at closed end. Antinode at open end
One closed end standing wave, fundamental frequency
1 node at closed end. Length is 0.25 of the wavelength. frequency = f0
One closed end standing wave, 2 nodes
1 node at closed end, one in middle. Length is 0.75 of the wavelength. frequency = 3*f0
One closed end standing wave, 3 nodes
1 node at closed end, 2 in middle. Length is 1.25 of wavelength. frequency = 5f0
Harmonics of fixed/2 open ends and 1 open end standing waves
Fixed/ 2 open: all integer multiples of fundamental frequency
1 open end: only odd integer multiples of fundamental frequency
To go up one octave…
double the frequency
Partial reflection superposistion
Transmitter + receiver with partial reflector, then reflector in front of it. Path difference 2x distance between partial reflector and reflector (d). If move partial reflector from one max to the next; path difference changed by 1 wavelength, so 2*change in d = 1 wavelength
Thin film interferance
Occurs as some light reflected of front off front of film, whilst other reflected off back. As light is emitted in short bursts, there is only coherency for a nano second; film just thin enough.
Young’s fringes equation
x=(lamda)*(L/d); x is fringe spacing; L is slit-screen distance; d is slit spacing. Can use to find lamda
Assumption made in Young’s fringes equation
sin(x) = tan(x) = x/L. Only works when angle is very very small
Diffraction gratings (multiple slits) equation
d*sin(theta) = n(lamda) at a max
Uses of diffraction gratings
Analysing light from stars; can see elements
Any sort of analysing light
Single slit diffraction equations (x2)
asin(theta) = n(lamda) at a MINIMA for n=1,2,3 but not 0. n=0 is the central maxima
W=lamda(L/a); where W is half width of central maxima; L is slit-screen distance; a is width of slit. Small angle assumption (so sin(x)=tan(x)) has been made