Chapter 8 Superposition

Question 1

The Principle of Superposition

Answer 1

states that when two or more waves with the same frequency travelling in opposite directions overlap, the resultant displacement is the sum of displacements of each wave

-The principle of superposition applies to all types of waves i.e. transverse and longitudinal

Question 2

The Principle of Superposition describes

Answer 2

how waves which meet at a point in space interact

Question 3

When two waves with the same frequency and amplitude arrive at a point, they superpose either:

Answer 3

• in phase, causing constructive interference. The peaks and troughs line up on both waves. The resultant wave has double the amplitude or, in anti-phase, causing destructive interference.
• The peaks on one wave line up with the troughs of the other. The resultant wave has no amplitude

Question 4

Stationary Waves

Answer 4

or standing waves, are produced by the superposition of two waves of the same frequency and amplitude travelling in opposite directions

-This is usually achieved by a travelling wave and its reflection. The superposition produces a wave pattern where the peaks and troughs do not move

—In order to produce a stationary wave, there must be a minima (node) at one end and a maxima (antinode) at the end with the loudspeaker

Question 5

Stretched strings

Answer 5

-Vibrations caused by stationary waves on a stretched string produce sound

—This is how stringed instruments, such as guitars or violins, work

-This can be demonstrated by a length of string under tension fixed at one end and vibrations made by an oscillator:

—As the frequency of the oscillator changes, standing waves with different numbers of minima (nodes) and maxima (antinodes) form

Question 6

Microwaves

Answer 6

• A microwave source is placed in line with a reflecting plate and a small detector between the two
• The reflector can be moved to and from the source to vary the stationary wave pattern formed -By moving the detector, it can pick up the minima (nodes) and maxima (antinodes) of the stationary wave pattern

Question 7

Air columns

Answer 7

-The formation of stationary waves inside an air column can be produced by sound waves

—This is how musical instruments, such as clarinets and organs, work

-This can be demonstrated by placing a fine powder inside the air column and a loudspeaker at the open end -At certain frequencies, the powder forms evenly spaced heaps along the tube, showing where there is zero disturbance as a result of the nodes of the stationary wave

Question 8

A stationary wave is made up

Answer 8

nodes and antinodes -The nodes and antinodes do not move along the string. Nodes are fixed and antinodes only move in the vertical direction -Between nodes, all points on the stationary are in phase

Question 9

nodes

Answer 9

are where there is no vibration

Question 10

antinodes

Answer 10

are where the vibrations are at their maximum amplitude

Question 11

fundamental mode of vibration or the first harmonic

Answer 11

When a stationary wave, such as a vibrating string, is fixed at both ends, the simplest wave pattern is a single loop made up of two nodes and an antinode

• This is called the fundamental mode of vibration or the first harmonic
• The particular frequencies (i.e. resonant frequencies) of standing waves possible in the string depend on its length L and its speed v

–As you increase the frequency, the higher harmonics begin to appear

-The frequencies can be calculated from the string length and wave equation

Question 12

Diagram showing the first three modes of vibration of a stretched string with corresponding frequencies

Answer 12

Question 13

One or two open ends in air column

Answer 13

Question 14

table showing all the different fundementials

Answer 14

Question 15

Diffraction

Answer 15

is the spreading out of waves when they pass an obstruction —-This obstruction is typically a narrow slit (an aperture) -The extent of diffraction depends on the width of the gap compared with the wavelength of the waves —-Diffraction is the most prominent when the width of the slit is approximately equal to the wavelength

Question 16

Diffraction is usually represented by a wavefront

Answer 16

as shown by the vertical lines in the diagram above -The only property of a wave that changes when its diffracted is its amplitude -This is because some energy is dissipated when a wave is diffracted through a gap

Question 17

Interference

Answer 17

Interference occurs when waves overlap and their resultant displacement is the sum of the displacement of each wave This result is based on the principle of superposition and the resultant waves may be smaller or larger than either of the two individual waves

Question 18

Interference of two waves can either be

Answer 18

-In phase, causing constructive interference. The peaks and troughs line up on both waves. The resultant wave has double the amplitude -In anti-phase, causing destructive interference. The peaks on one wave line up with the troughs of the other. The resultant wave has no amplitude

Question 19

Waves are coherent if they have

Answer 19

the same frequency and constant phase difference

Question 20

Coherence is vital

Answer 20

is vital in order to produce an observable interference pattern

Question 21

Demonstrating Two Source Interference: Using Water Waves

Answer 21

• Two-source interference in can be demonstrated in water using ripple tanks
• The diagram below shows diffracted circle shaped water waves from two point sources eg. dropping two pebbles near to each other in a pond
• The two waves interfere causing areas of constructive and destructive interference
• The lines of maximum displacement occur when all the peaks and troughs line up with those on another wave

Question 22

Demonstrating Two Source Interference: Using Sound Waves

Answer 22

• Two source interference for sound waves looks very similar to water waves
• Sound waves are longitudinal waves so are made up compressions and rarefactions
• Constructive interference occurs when the compression and rarefactions line up and the sound appears louder -Destructive interference occurs when the compression lines up with a rarefaction and vice versa. The sound is quieter

—–This is the technology used in noise cancelling headphones

Question 23

Demonstrating Two Source Interference: Using Microwaves

Answer 23

• Two source interference for microwaves can be detected with a moveable microwave detector
• Constructive interference: regions where the detector picks up a maximum amplitude
• Destructive interference: regions where the detector picks up no signal

Question 24

Demonstrating Two Source Interference: Using Light Waves

Answer 24

• For light rays, such as a laser light through two slits, an interference pattern forms on the screen
• Constructive interference is shown as bright fringes on the screen The highest intensity is in the middle
• Destructive interference is shown as the dark fringes on the screen These have zero intensity

Question 25

For two-source interference fringes to be observed, the sources of the wave must be:

Answer 25

• Coherent (constant phase difference)
• Monochromatic (single wavelength)
• When two waves interfere, the resultant wave depends on the phase difference between the two waves
• This is proportional to the path difference between the waves which can be written in terms of the wavelength λ of the wave

Question 26

When two waves interfere, the resultant wave depends on the

Answer 26

phase difference between the two waves -This is proportional to the path difference between the waves which can be written in terms of the wavelength λ of the wave

-The difference in distance is the path difference

Question 27

For constructive interference (or maxima)

Answer 27

-the difference in wavelengths will be an integer number of whole wavelengths

—n is the order of the maxima/minima since there is usually more than one of these produced by the interference pattern

-path difference equation= nλ

Question 28

For destructive interference (or minima)

Answer 28

• it will be an integer number of whole wavelengths plus a half wavelength

• n is the order of the maxima/minima since there is usually more than one of these produced by the interference pattern
• path difference equation = (n+0.5)λ

Question 29

Double Slit Interference

Answer 29

• experiment demonstrates how light waves produced an interference pattern
• When a monochromatic light source is placed behind a single slit, the light is diffracted producing two light sources at the double slits A and B
• Since both light sources originate from the same primary source, they are coherent and will therefore create an observable interference pattern
• Both diffracted light from the double slits create an interference pattern made up of bright and dark fringes
• The wavelength of the light can be calculated from the interference pattern and experiment set up.

Question 30

These are related using the double-slit equation

Answer 30

Question 31

The interference pattern on a screen will show as ‘fringes’ which are dark or bright bands

Answer 31

• Constructive interference is shown through bright fringes with varying intensity (most intense in the middle)
• Destructive interference is shown from dark fringes where no light is seen -A monochromatic light source makes these fringes clearer and the distance between fringes is very small due to the short wavelength of visible light

Question 32

The Grating Equation

Answer 32

Question 33

A diffraction grating

Answer 33

is a plate on which there is a very large number of parallel, identical, close-spaced slits

-When monochromatic light is incident on a grating, a pattern of narrow bright fringes is produced on a screen

Question 34

Angular Separation

Answer 34

• The angular separation of each maxima is calculated by rearranging the grating equation to make θ the subject
• The angle θ is taken from the centre meaning the higher orders are at greater angles
• The angular separation between two angles is found by subtracting the smaller angle from the larger one -The angular separation between the first and second maxima n1 and n2 is θ2 – θ1
• The maximum angle to see orders of maxima is when the beam is at right angles to the diffraction grating. This means θ = 90o and sin(θ) = 1

Question 35

Determining the Wavelength of Light: Improving experiment and reducing uncertainties

Answer 35

• The fringe spacing can be subjective depending on its intensity on the screen.
• Take multiple measurements of h (between 3-8) and finding the average
• Use a Vernier scale to record h, in order to reduce percentage uncertainty
• Reduce the uncertainty in h by measuring across all fringes and dividing by the number of fringes
• Increase the grating to screen distance D to increase the fringe separation (although this may decrease the intensity of light reaching the screen)
• Conduct the experiment in a darkened room, so the fringes are clearer -Use grating with more lines per mm, so values of h are greater to lower percentage uncertainty

Question 36

Determining the Wavelength of Light: Method

Answer 36

• The wavelength of light can be determined by rearranging the grating equation to make the wavelength λ the subject
• The value of θ, the angle to the specific order of maximum measured from the centre, can be calculated through trigonometry
• The distance from the grating to the screen is marked as D -The distance between the centre and the order of maxima (e.g. n = 2 in the diagram) on the screen is labelled as h – the fringe spacing
• Measure both these values with a ruler -This makes a right-angled triangle with the angle θ as the ratio of the h/D = tanθ

Question 37

The wavelength of light is calculated by the angle to the order of maximum

Answer 37