Chapter 6

Question 1

Amplitude

Answer 1

Max displacement from the undisturbed postion

Question 2

Wavelength

Answer 2

Distance between two adjacent points that are in phase

Question 3

Time period

Answer 3

Time to complete one wave cycle

Question 4

Frequency

Answer 4

Number of cycles or oscillations per second. Number of wave crests passing a point per second

Question 5

frequency =

Answer 5

1/Time period

Question 6

Wave speed =

Answer 6

frequency * wavelength

Question 7

Speed of sound (air)

Answer 7

330 m/s

Question 8

Speed of light (vacuum)

Answer 8

3.0*10^8 m/s

Question 9

Phasors

Answer 9

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.

Question 10

General wave equation

vertical displacement, y =

Answer 10

Asin(2pift) (A is amplitude; 2pi is radians; f is frequency; t is time passed)
or Asin(w*t); w is angular velocity

Question 11

Angular velocity, w (omega) =

Answer 11

2pif

Question 12

Superposition

Answer 12

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

Question 13

In phase

Answer 13

Phasors at same angle

Question 14

Anti phase

Answer 14

Phasors at 180(deg) or 2*pi(rad)

Question 15

Constructive interferance

Answer 15

Occurs when 2 waves are in phase; phasors at same angle; maximum is heard; path difference is n(lamda)

Question 16

Destructive interferance

Answer 16

Occurs when 2 waves are in anti phase; phasors at 180(deg)/2pi(rad) ; minimum is observed; path difference is (2n+10.5(lamda)

Question 17

To observe a stable interference pattern

Answer 17

Waves must be coherent

Question 18

Coherent

Answer 18

Waves have a constant phase difference

Question 19

Phase change on reflection

Answer 19

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

Question 20

Standing wave

Answer 20

Superposition pattern of 2 identical waves, travelling in opposite directions

Question 21

Identical waves

Answer 21

Same frequency, wavelength and speed; ideally same amplitude

Question 22

Distance between adjacent nodes (or antinodes) on a standing wave

Answer 22

1/2(lamda)

Question 23

Where are standing waves most clear

Answer 23

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

Question 24

Lloyd’s mirror

Answer 24

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

Question 25

Huygen’s theory

Answer 25

Every point on a wave is a secondary source, so multiple wave lengths are emitted. The next wavefront is their superposition pattern

Question 26

Fixed end standing waves, fundamental frequency

Answer 26

2 nodes, each at fixed end. Length, L is half the wavelength. frequency = f0

Question 27

Fixed end standing waves, 3 nodes

Answer 27

2 nodes at each end, one in middle. Length is the wavelength. frequency = 2*f0

Question 28

Fixed end standing waves, 4 nodes

Answer 28

2 nodes at each end, 2 in middle. Length is 1.5 wavelengths. frequency = 3*f0

Question 29

Fixed end standing waves, 5 nodes

Answer 29

2 nodes at each end, 3 in middle. Length is 2 wavelengths. Frequency = 4*f0

Question 30

One closed end standing wave

Answer 30

1 node at closed end. Antinode at open end

Question 31

One closed end standing wave, fundamental frequency

Answer 31

1 node at closed end. Length is 0.25 of the wavelength. frequency = f0

Question 32

One closed end standing wave, 2 nodes

Answer 32

1 node at closed end, one in middle. Length is 0.75 of the wavelength. frequency = 3*f0

Question 33

One closed end standing wave, 3 nodes

Answer 33

1 node at closed end, 2 in middle. Length is 1.25 of wavelength. frequency = 5f0

Question 34

Harmonics of fixed/2 open ends and 1 open end standing waves

Answer 34

Fixed/ 2 open: all integer multiples of fundamental frequency
1 open end: only odd integer multiples of fundamental frequency

Question 35

To go up one octave…

Answer 35

double the frequency

Question 36

Partial reflection superposistion

Answer 36

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

Question 37

Thin film interferance

Answer 37

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.

Question 38

Young’s fringes equation

Answer 38

x=(lamda)*(L/d); x is fringe spacing; L is slit-screen distance; d is slit spacing. Can use to find lamda

Question 39

Assumption made in Young’s fringes equation

Answer 39

sin(x) = tan(x) = x/L. Only works when angle is very very small

Question 40

Diffraction gratings (multiple slits) equation

Answer 40

d*sin(theta) = n(lamda) at a max

Question 41

Uses of diffraction gratings

Answer 41

Analysing light from stars; can see elements

Any sort of analysing light

Question 42

Single slit diffraction equations (x2)

Answer 42

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