Superposition Summary notes

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Superposition

1. Diffraction

Diffraction is the spreading of waves when they pass through an opening or around an obstacle.

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2. Principle of superposition

If two or more waves of the same kind meet at a point, the resultant displacement is the vector sum of the individual displacements due to the waves at the point.

3. Coherence

Two wave sources are said to be coherent if they have a constant phase difference. This necessarily means that the two waves are of the same frequency.

4. Path difference

The difference between the distances traveled by the two waves from the sources to the point where they meet. 5. Interference

Interference occurs when two or more waves of the same type superpose at a point in space to give a resultant wave whose amplitude is given by the principle of superposition.

Constructive interference occurs when the two waves meet in phase, or the path difference is an integer number of wavelengths, i.e. path difference = nλ and phase difference _ _

 

n 2 2n

Destructive interference occurs when the two waves meet in antiphase, or the path difference is an odd number of half wavelengths, i.e. path difference = (2n + 1) λ/2 and phase difference _ _

   2 2 (2 1) ) 1 2 (      n n

6. For a two-source interference to be observable, the conditions required are that the two waves

(i) must be coherent. This is to ensure that their phase relationship does not change with time so that our eyes can observe a ‘stationary’ pattern.

(ii) must have roughly the same amplitude. This is to ensure good contrast between maxima and minima of intensity. (iii) for transverse waves, must be completely unpolarised or polarized in the same plane.

7. Interference of water waves in a ripple tank

All points on AB are equidistant from S1and S2. The path difference is zero and the two waves meet along AB in phase. At all points along EF the path difference is one wavelength, and the two waves also meet in phase. There is constructive interference. Lines AB and EF are called antinodal lines.

Points on CD are half a wavelength nearer to S1than S2. The path difference is half a wavelength. The crests (or troughs) from S1arrive simultaneously with troughs (or crests) from S2. The waves arrive in antiphase and there is destructive interference. Similarly, there is destructive interference along GH where the path difference is 1½ wavelengths. Lines CD and GH are called nodal lines.

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8. Interference of light

Monochromatic light falls on single slit S (which acts as a point light source). Diffraction at this slit causes light to spread and fall on the two slits S1and S2(which act as coherent sources). Superposition occurs in the area where the diffracted beams from S1and S2overlap. Alternate bright and dark equally spaced fringes can be observed at the screen. The intensity pattern is as shown below. Note that we usually assume that the fringes are of equal brightness.

Fringe separation

a D

x , where λ is the wavelength of the light, D is the distance between the double slits and the

screen and a is the separation of the double slits.

Note that if more than one wavelength is used, each wavelength will form its own fringe pattern and the patterns will overlap one another.

When certain parameters are changed, you may be asked to comment on the change to the fringe separation, intensity etc.

9. Diffraction grating

A diffraction grating has many closely spaced slits. When a beam of light incidents on a diffraction grating, each slit acts as a coherent source. Interference of the waves generated from each slit produces bright lines in a certain directions. In these directions, all the waves interfere in phase. Reinforcement of the waves from each slit occurs and constructive interference occurs.

The relevant formula is d sin θ = n λ, where d is the separation between the slits and n is the order of the fringe. Note that d = 1/N where N is the number of slits per unit length.

10. Stationary waves

A stationary or standing wave is produced when two progressive waves of equal frequency and amplitude traveling with the same speed in opposite directions superpose. Positions along the wave that are

permanently at rest (zero displacement) are called nodes, while positions along the wave that have the greatest amplitude of vibration are called antinodes.

t = T/8 t = T/4 t = 3T/8 t = T/2 t = T/2 t = 5T/8 t = 3T/4 t = 7T/8 t = T

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11. Comparison between progressive and stationary waves

Stationary wave Progressive wave

Amplitude Varies according to position from zero at the nodes to a maximum at 2a at the antinodes

same for all particles in the path of the waves

Frequency All particles vibrate in SHM with the same frequency

(except for those at the nodes which are at rest) All particles vibrate in SHM with the frequency of the wave. Wavelength It is equal to twice the distance between two

adjacent nodes or two adjacent antinodes

It is equal to the distance between adjacent particles which have the same phase Phase Phase of all particles between two adjacent nodes is

the same.

Particles in adjacent segments have a phase difference of  rad

All particles within one wavelength have different phases.

Waveform Does not advance. Advances with the velocity of the wave.

Energy No translation of energy, but there is energy stored in the wave.

Energy translation in the direction of travel of the wave.

12. Experimental set-up to demonstrate stationary waves for microwaves

Distance between two consecutive minima (nodes) is equal to half the wavelength of the microwaves.

13. Experimental set-up to demonstrate stationary waves in a stretched string

The speed of waves in the string is determined by the mass per unit length of the string and the tension in the string. The string is made (driven) to vibrate at the frequency of the signal generator. The string would vibrate with greatest amplitude in several frequencies (its natural modes of

vibration). For the stationary wave that is formed, the two fixed ends must be nodes because the string there is not free to vibrate. The first few modes of vibration are shown below. f1is called the fundamental frequency.

To signal generator f1= v/λ1= v/2L f2= v/λ2= v/L = 2 f1 f3= v/λ3= 3v/2L = 3 f1 f4= v/λ4= 2v/L = 4 f1

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14. Experimental set-up to determine stationary waves for air columns

The closed end of a column is a displacement node where the particles cannot vibrate. The open end of a column is a displacement antinode because the air there is free to vibrate.

Closed tube

Let the fundamental frequency be f1.

f1= v/λ1= v/4L, where v is the speed of the air in the tube.

This is the first overtone. f2= v/λ2= 3v/4L = 3 f1

This is the second overtone. f3= v/λ3= 5v/4L = 5 f1

The frequencies of vibration in a closed tube are odd multiples of the fundamental frequency. Open tube

Let the fundamental frequency be f1. f1= v/λ1= v/2L

Verify for yourselves that the frequencies of vibration in an open tube are integer multiples of the fundamental frequency. 15. Measurement of wavelength of sound using stationary waves through the resonance air tube method

 Choose a frequency.

 Adjust the length of the tube until there is resonance – a loud sound will be heard.

 Adjust the length until the next loud sound is heard.  The difference in length is equal to half a wavelength,

i.e. λ = 2 (L2– L1).

 Obtain the speed of sound from v = fλ.

Note that in this experiment, stationary waves are set up with a displacement node at the water surface and a displacement antinode at the open end of the tube.

L = ¼1

L = ¾ 2

L = 5/4 3

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Superposition Discussion Questions

1. Two loudspeakers S

1

and S

2

each emit sound of frequency 330 Hz uniformly in all directions. S

1

has an acoustic output of 1.2 x 10

-3

_{W while S}

2

has an acoustic output of 1.8 x 10

-3

W. S

1

and S

2

vibrate in phase. A point P is 4 m from S

1

and 3 m from S

2

. The speed of sound in air is 330 m s

-1

.

(a) How are the phases of the two waves arriving at P related?

(b) What is the intensity of sound at P with

(1) S

1

on alone,

(2) S

2

on alone, and

(3) S

1

and S

2

both on. [5.97 x 10

-6

W m

-2

, 1.6 x 10

-5

W m

-2

, 4.15 x 10

-5

W m

-2

]

S1

S2

P

4 m

3 m

a) Phase Difference =





1

2

3

4

2 









x

rad

 The two waves are in phase

b) (1) S

1

alone:

₂ 6 3 1 1 1

5 .

97

10 )

4 (

4

10

2 .

1 



_

_







A

P

I

Wm

-2

(2) S

2

alone:

₂ 5 3 2 2 2

1 .

59

10 )

3 (

4

10

8 .

1 



_

_







A

P

I

Wm

-2

(3) S

1

and S

2

?

We cannot add intensity, but can only add amplitude. This is because adding of intensity

itself does NOT take into consideration the effects of superposition.

2 1 1

a

I



I

₁



ka

₁2

k

I

a

1 1



2 2 2

a

I



I

₂



ka

₂2

k

I

a

2 2



a

resultant

= a

1

+ a

2

=

k

I

1



2

I

resultant

= ka

resultant2

=

5 2 1 2 1 2 2 1

10

14 .

4

2 )

(





I



I



I







k

I

k

Wm

-2