Waves - Level 3 Flashcards

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1
Q

Harmonic frequencies

A

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

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2
Q

Fundamental frequency/first harmonic

A

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).

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3
Q

Harmonics in a column with both ends open

A

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)

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4
Q

Why are other frequencies produced when a whistle is blown?

A

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.

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5
Q

Beat frequency

A

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.

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6
Q

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

A

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
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7
Q

Harmonics in a column with both ends closed

A

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)

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8
Q

Calculate the frequency of any harmonic

A
  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)
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9
Q

Relationship between pulse speed, tension and linear density

A

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

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10
Q

Interference

A

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

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11
Q

Interference conditions for light /sound sources

A
  1. Light sources must be coherent, maintaining a constant phase relationship
  2. Light must be monochromatic, consisting of just one wavelength (λ)
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12
Q

Principle of superposition of waves

A

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.

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13
Q

Constructive interference

A

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.

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14
Q

Deconstructive interference

A

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.

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15
Q

Path difference and phase difference of interfering waves

A

(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…

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16
Q

Path difference and phase difference of interfering waves

A

(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…

17
Q

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

A

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

18
Q

Assumptions for Young’s double slit experiment

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

- The two light rays are parallel at the start

19
Q

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

A

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

20
Q

Diffraction grating

A

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.

21
Q

Calculate the distance between the two wave sources

A

d = D/N

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

22
Q

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

A

d y/L = n λ

23
Q

White light

A

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.

24
Q

Overlapping of colours

A

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

25
Q

How can the same colour have a different frequency?

A

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.
26
Q

Formula for the apparent (not real) frequency

A

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)
27
Q

Red shift - spectra from distant galaxies

A

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’.

28
Q

Effect of red shift

A

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

29
Q

Red shift in galaxies

A

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.

30
Q

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

A

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

31
Q

What determines Doppler shift?

A

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

32
Q

Doppler effect

A
  • 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