Waves - Level 3

Question 1

Harmonic frequencies

Answer 1

Standing wave patterns are only created within a medium at specific frequencies of vibration. These frequencies are known as harmonic frequencies, (harmonics).

Question 2

Fundamental frequency/first harmonic

Answer 2

The fundamental frequency/1st harmonic is the lowest possible frequency at which a string could vibrate to form a standing wave pattern (only 1 loop is formed).

Question 3

Harmonics in a column with both ends open

Answer 3

The standing wave patterns will have anti-nodes at each end of the air column and a node in between.

λ = 2L
f = v/λ

(>< shape)

Question 4

Why are other frequencies produced when a whistle is blown?

Answer 4

The other frequencies produced are overtones (frequencies that are higher than the fundamental frequency of the vibration). When the whistle is blown, a mixture of overtones produces a distinctive sound (timbre). This causes different tones/beats to be produced between fundamental frequencies.

Question 5

Beat frequency

Answer 5

A beat is an interference between two sound waves of slightly different frequencies. When two sound waves of slightly different frequencies approach your ear, the alternating constructive and destructive interference produces beats, causing the sound to be alternatively soft and loud. The rate of beats is the differencebetween the two frequencies.

Question 6

Harmonics in a column with one end open, one end closed

Answer 6

The standing wave patterns will have a node at one end and an anti-node at the other end of the air column.

λ = 2L
f = v/λ

(< shape)

• Only odd harmonics of the frequencies are produced

Question 7

Harmonics in a column with both ends closed

Answer 7

The standing wave patterns will have nodes at each end of the air coluumn and an anti-node in between.

λ = 2L
f = v/λ

(<> shape)

Question 8

Calculate the frequency of any harmonic

Answer 8

1. Calculate the frequency of the 1st harmonic
2. Multiply the frequency value by the number of the harmonic you want to calculate
   (f1 x 2 = f2)

Question 9

Relationship between pulse speed, tension and linear density

Answer 9

The velocity of the wave depends on the linear density (µ) of and the tension (T) in the string. Making the string tighter and lighter increases the pulse speed, and making the string loser and heavier slows the pulse speed.

v = τµ

Question 10

Interference

Answer 10

Interference is the combination of two or more waves to form a resultant wave.

Question 11

Interference conditions for light /sound sources

Answer 11

1. Light sources must be coherent, maintaining a constant phase relationship
2. Light must be monochromatic, consisting of just one wavelength (λ)

Question 12

Principle of superposition of waves

Answer 12

When two or more propagating waves of same
type are incident on the same point, the total displacement at that point is equal to thevector sumof
the displacements of the individual waves.

Question 13

Constructive interference

Answer 13

Constructive interference is when acrestof a wave meets a crest of another wave, or a troughof a wave meets a trough of another wave, having the same
frequency at the same point. The total displacement of the waves is equal to the sum of the individual displacements.

Question 14

Deconstructive interference

Answer 14

Deconstructive interference is when a crest of one wave meets a trough of another wave. The total
displacement of the waves is equal to the difference between the individual displacements.

Question 15

Path difference and phase difference of interfering waves

Answer 15

(Antinodal lines - maximum points - constructive interference)

From the centre anti-nodal point onwards
Path difference - 0, 1λ, 2λ, 3λ…nλ

Phase difference - 0, 2π, 4π, 6π…

Question 16

Path difference and phase difference of interfering waves

Answer 16

(Nodal lines - deconstructive interference)

From the nodal point next to the the centre onwards
Path difference - λ/2, 3 λ/2, 5 λ/2…

Phase difference is π, 3π, 5π, 7π…

Question 17

Calculating bright (anti-nodal) and dark (nodal) fringes will appear for a larger angle

Answer 17

d Sin θ = path difference

• For bright fringes, path difference is n λ
• For dark fringes, path difference is (n - 1/2) λ

Where,
d is the distance between the two wave sources
θ is the angle between the central anti-nodal line and line from centre to nth point
n is the count of how many bright fringes there are from thecentral fringe

Question 18

Assumptions for Young’s double slit experiment

Answer 18

• The path difference (d) and angle (θ) is very small

- The two light rays are parallel at the start

Question 19

Calculate the angle between the central anti-nodal line and line from centre to nth point

Answer 19

To calculate the angle (θ), use Tan θ = y/L

Where,
y is the distance between the central anti-nodal line and the nth point
L is the length of the central anti-nodal line

Question 20

Diffraction grating

Answer 20

A diffraction grate (glass plate), contains alarge number of parallel, closely spaced slits. When light encounters an entire array of identical, equally-spaced slits,the bright fringes, which come from constructive interference of the light waves from different slits, are formed because more light is coming from many slits, and so the intensity of the light becomes greater.

Question 21

Calculate the distance between the two wave sources

Answer 21

d = D/N

Where,
D is the length of the grating
N is the number of slits on the grating

Question 22

Calculating bright (anti-nodal) and dark (nodal) fringes will appear for a smaller angle

Answer 22

d y/L = n λ

Question 23

White light

Answer 23

Sin θ increases with wavelength λ, red light which has the longest wavelength is diffracted through the largest angle. Violet light has the shortest wavelength and is
diffracted the least. Thus, white light is split into its component colours from violet
to red light. The spectrum is repeated in the different orders of diffraction.

Question 24

Overlapping of colours

Answer 24

Two colours of different orders may overlap if their angles of diffraction θ are equal.
Since d and θ are the same, the condition for overlapping of spectra of two different colours is
n1λ1 = n2 λ2

Question 25

How can the same colour have a different frequency?

Answer 25

When an object reflecting light moves relative to an observer.

• When approaching, the observed frequency is greater than that being emitted in receding.
• We perceive this frequency shift as a change in the colour of the object.
• A higher frequency (object approaching) corresponds to a more blue colour, a lower frequency (object receding) to a more red colour.

Question 26

Formula for the apparent (not real) frequency

Answer 26

f’ = fo (Vw / Vw±Vs)

Where, 
f' is the apparent frequency 
fo is the actual frequency 
Vw is the wave velocity in the given medium (in air)
Vs is the velocity of the moving object 

• Use a positive sign if the object is moving away from the observer (λ increases, f decreases)
• Use a negative sign if the object is moving towards the observer (λ decreases, f increases)

Question 27

Red shift - spectra from distant galaxies

Answer 27

Red shift occurs because our Sun contains helium. We know this because there are black lines in the spectrum of the light from the Sun where helium has absorbed light. These lines form the absorption spectrum for helium.

When we look at the spectrum of an element like helium of a distant star, we still see an
absorption spectrum of similar pattern. However, the pattern of lines has moved towards
the red end of the spectrum, which is ‘red shift’.

Question 28

Effect of red shift

Answer 28

Wavelength increases up towards the red and beyond (frequency decreases). The positions of the lines have changed because of the Doppler effect. dsinθ = nλ

Question 29

Red shift in galaxies

Answer 29

For the red line angle θ must have increased as the redline has shifted out from the centre. This indicates that the wavelengths of the light have increased and their frequencies have decreased. According to
Doppler Effect the star (source) must be moving away from the lab/earth. Astronomers have found that the further from us a star is, the more its light is red-shifted. This tells us that distant galaxies are moving away from us, and that the further away a galaxy is, the faster its moving away.

Question 30

How are standing waves produced in a column with one end open, one end closed

Answer 30

Sound waves enter at the open end, travel
along the pipe and are reflected from the closed
end.
- Reflected waves are 180 degreesout of phase
making the closed-end a place of permanent
destructive interference (a node).
- Reflected waves of the correct wavelength reflect from
the open end in phase with incident waves,
producing a position of permanent
constructive interference (an antinode) .
- The amplitude at the antinode is larger than
the amplitude of the wave

Question 31

What determines Doppler shift?

Answer 31

The relative velocity between two objects determines the Doppler-shifted frequency perceived.

Question 32

Doppler effect

Answer 32

• Compared to the emitted frequency, the received frequency is higher during the approach, identical at the instant of passing by, and lower during the recession.

If moving object accelerates away from other object

• The distance travelled by the object between the creation of each successive wave front is increasing, (relative velocity of the object is increasing, so ​the actual and apparent wavelength is increasing) but the velocity of sound in air remains constant.
• The relative velocity between two objects determines the Doppler-shift frequency perceived, and because this is increasing, the frequency is decreasing.

VICE VERSA for if moving object accelerates toward other object