Chapter 16: Diffraction solutions
1. [4 marks] (VCE 2001 Q11)
When the wavelength is long the sound waves can diffract around the head, so the intensity for both ears will be roughly the same. The sound takes a little of time to travel to the other ear, the brain can detect the time difference between the compressions reaching the first ear then the other. Using this time difference the brain calculates where the sound is coming from.
Short wavelengths
This does not work as well when the wavelength is smaller, the time difference is too small and the brain cannot detect the difference as well. The way we overcome this is that the smaller wavelengths do not diffract around the head as well, and therefore the intensity on one side of the head is more intense than the other. The brain uses this to determine where the sound is coming from.
2. [2 marks] (VCE 2002 Q6)
The amount of diffraction depends upon the wavelength of the sound compared with the width of the door.
This is expressed as the ratio where w is the width of the door and diffraction is the most when this ratio is > 1, ie when l > 1m. Since v = fl (the wave equation) and
v = 340m/s, the corresponding frequency is 340Hz.
An estimate of what frequencies will have a reduced intensity is when the ratio is less than 1.
This will result in all frequencies above 340Hz diffracting less to some extent and so won’t be heard as loudly by Peta.
\ 340 – 22000Hz (ANS)
3. [2 marks] (VCE 2003 Q7)
Use v = fl
\ 340 = 6000 ´ l
\ l =
= 5.7cm (ANS)
(Make sure you round off correctly)
4. [2 marks] (VCE 2003 Q8)
The amount of diffraction depends upon the wavelength of the sound compared with the width of the speaker.
This is expressed as the ratio where d is the width of the speaker and diffraction is the most when this ratio is > 1.
At 6000Hz the wave length is 5.7cm. This means that not much diffraction will occur, and so this sound will be very directional, and only project where the actual speaker is pointing. This means that Mustafa won’t hear this frequency very well.
Rebus will hear all the frequencies being produced, but only the lower frequencies will diffract sufficiently for Mustafa to hear.
5. [3 marks] (VCE 2003 Q9)
Sound can’t travel directly to the houses, so it must get there through transmission, diffraction and reflection.
Barrier B absorbed and reflected sound, thus reducing the noise level at all houses. Some sound would diffract over the barrier to reach all houses, but more would reach house 3 because it does not have to diffract as much. Sound would also reflect from barrier A to reach house 3.
(You needed to make the last comment to get full marks for this question.)
6. [2 marks] (VCE 2003 Q10)
The sound reflected by barrier A was now scattered (diffuse reflection) by the rough wall, so a lot less sound was reflected. Some extra sound was also absorbed by the wall. This will have much more effect on house 3, as this as the one that had most of the reflected sound getting to it.
7. [4 marks] (VCE 2004 Q7)
The graph of amplitude arises as a result of diffraction of the sound passing through the gap.
Diffraction means that the sound spreads out after passing through the gap. The amount of diffraction depends upon the ratio of the wavelength to the gap size. Since there is a lot of diffraction then
~ 1.
The gap size is 0.30 m and so wavelengths greater than 0.30 m will show significant diffraction. Given that the speed of sound is 340ms-1 then a wavelength of 0.30 m corresponds to a frequency of 1133Hz.
Hence, the approximate frequency used was about 1000 Hz.
(Answers in the range 500 – 1500 Hz were accepted.)
8. [2 marks] (VCE 2004 Q8)
The curve shows very little diffraction and so the wavelength is considerably less than 0.30m. Correspondingly the frequency will be considerably higher than 1000 Hz, as calculated in question 549. Thus, the frequency of 10 000 Hz as listed in A is best fits with Lee’s prediction.
\ A (ANS)
9. [2 marks] (VCE 2004 Sample Q7)
D (ANS)
10. [3 marks] (VCE 2004 Sample Q8)
The amount of diffraction depends upon the wavelength of the sound compared with the width of the speaker.
This is expressed as the ratio where w is the width of the speaker and diffraction is the most when this ratio is > 1. The higher frequencies with their smaller wavelengths don’t diffract enough for Leo to hear them.
Louise will hear all the frequencies because she is directly in front of the speaker. She will hear all the frequencies produced by the speaker, in the correct proportions.
11. [4 marks] (VCE 2004 Pilot Q10)
a) Use λ =
= 3.4 m (ANS)
b) The low frequency sound (100 Hz) with a wavelength of 3.4m will diffract around the screen because the screen is smaller than the wavelength. These sounds will be heard well.
The higher frequencies of the dialogue (3000 Hz) will have λ around 0.1 m and so will not diffract around the screen. These sounds will not be heard well
12. [3 marks] (VCE 2006 Q1)
Jamie is listening to the sound of an orchestra through a small gap in a partly open sliding door. When the sound wave travels through the gap, [diffraction] occurs and spreading of the wave results. High pitched (frequency) instruments such as flutes experience [less] spreading than lower pitched instruments. As the size of the gap decreases, the angle of spreading will [increase].
13. [3 marks] (VCE 2007 Q8)
60 = 10 log
where I0 = 1 x 10-12
\ 6.0 = log
\ = 106.0
\ I = 106.0 ´ 1 ´ 10-12
\ I = 10-6.0
I = 1 ´ 10-6 W m-2 (ANS)
14. [1 + 2 = 3 marks] (VCE 2007 Q9)
a) 200 Hz (ANS)
b) To reach the point Y, the sound will need to diffract. The amount of diffraction depends upon the wavelength of the sound compared with the width of the stage opening (w)
This is expressed as the ratio
The higher frequencies with their smaller wavelengths don’t diffract as much and so will sound softer at the point Y.
15. [2 marks] (VCE 2007 Q10)
The sound level has dropped by 9dB. Each drop of 3dB corresponds to a halving of the intensity. Therefore the intensity has halved three times, so the ratio is 8:1
\ C (ANS)
16. [2 marks] (VCE 2007 Q11)
Doubling the distance will decrease the sound intensity (Wm-2) by a factor of 4. This leads to a 6dB drop in sound level.
\ B (ANS)
17. [2 marks] (VCE 2008 Q12)
The gap on the stage is approximately 3.0m, sound with of more than 100 Hz will not diffract as much as those below this value. This means that the outer edges of the first few rows will not hear the sound at the same level as those people sitting towards the middle of the room. Thus the sound will be distorted for the people in the outer edges of the first few rows due to diffraction the lower frequencies will be louder to these people compared to the higher frequencies.
\ B (ANS)
18. [2 marks] (VCE 2008 Q13)
The sound will reflect off the walls in many different directions and this will lead to many paths for the sound to reach people from the one speaker. This will then lead to interference in some parts of the theatre.
\ D (ANS)