Chapter 12: Sound concepts solutions
1. [2 marks] (VCE 2001 Q1)
f =
The period is the time taken to complete one cycle.
\ f =
\ f = 200Hz (ANS)
2. [2 marks] (VCE 2001 Q2)
Using v = fl
\ l=
\ l =
\ l = 1.7 m (ANS)
3. [2 marks] (VCE 2001 Q3)
The time delay was given by
= 0.005 secs.
Starting with elimination of the incorrect answers. B and D were out with the timing. The distance of the microphone is 1.7 m away from the speaker, therefore a full wave would fit inside the gap. (or it won’t start reading until 0.005 seconds) . The speaker would normally start at its mean or central position and produce either a compression or rarefaction as its first movement.
\ A (ANS)
4. [4 marks] (VCE 2001 Q4)
Divide your answer into two parts:
Why sound persisted and High and Low frequency
Why sound persisted.
· The reason the sound was heard for some time afterwards was that the sound was reflecting off all the walls and ceiling.
· The stone walls and ceiling (all hard surfaces) allow for a greater amount of the sound to be reflected. Therefore it remains in the cathedral for a longer period of time, as it travels a greater distance.
· As the walls, ceiling and floor absorb the sound energy the intensity drops.
High and Low frequency.
· Looking at the graph of the human ear over a frequency range, people are more sensitive to the middle frequency (~ 3000 Hz).
· This means that as the intensity drops off the high and low frequencies can no longer be detected by the human ear. (or alternatively, the mid-range can be heard at lower intensities).
· So Mel and Jill will no longer be able to hear the sounds at the high and low frequencies.
5. [2 marks] (VCE 2001 Q5)
The frequency that they would hear longest is the frequency that is the lowest point on the graph.
The sound is going to decay with time, so the lowest point on the graph is the frequency that the ear will hear when it is softest. (i.e. after the most decay)
\ 3500 Hz (ANS)
6. [2 marks] (VCE 2001 Q8)
\ R2 = 20 m (ANS)
Alternatively, the new intensity is of the original, so the distance will be twice the original. \ 20m
7. [2 marks] (VCE 2001 Q9)
(ANS)
Alternatively, if the distance has doubled then the intensity must change by a factor =.
Each change of means a -3dB change
\ a change of will be a –6dB change.
8. [2 marks] (VCE 2002 Q1)
As Ashley is beating the drum at 2 beats per second, the time between beats is 0.5 secs.
If Pat hears the drum at the same instant that the drum is being hit, then he must be the distance away that the sound travels in 0.5 secs.
\ Speed of sound v =
=
= 334 m/s (ANS)
9. [4 marks] (VCE 2002 Q2)
Morgan is more correct.
This is because sound is a wave, and there is no net displacement of the medium. The dust particle will only move back and forwards about its mean position.
10. [2 marks] (VCE 2002 Q3)
After T/4, each point on the wave will have moved ¼ of a cycle to the left. \ E
After T/2, each point on the wave will have moved ½ of a cycle to the left. \ C
T/4 = E T/2 = C (ANS)
11. [2 marks] (VCE 2002 Q4)
If the diagram corresponds to a standing wave, then it is showing nodes and antinodes. The nodes remain stationary, and the antinodes vary from positive maximum through to negative maximum and back.
\ ¼ of a cycle later the maximum crests have moved to the zero line. \ D
After ½ of a cycle, the positive maximum has moved to show a negative maximum, etc. \ C
T/4 = D T/2 = C (ANS)
12. [2 marks] (VCE 2002 Q7)
Read off the graph by moving up the y-axis until the value of 10-5 Wm-2 is reached and then horizontally to read off the values of the graph that correspond:
Lowest Hz 40Hz Highest Hz 20 000 Hz (ANS)
13. [2 marks] (VCE 2002 Q8)
Use the intensity level formula
So the level is reduced by 40dB
14. [2 marks] (VCE 2002 Q9)
This is based on the inverse square relationship between distance and intensity.
The distance is doubled and therefore the intensity changes by a factor of
Hence response A (ANS)
15. [2 marks] (VCE 2003 Q1)
Sound is a longitudinal wave. The speaker created pressure variations which travelled away from the speaker as a longitudinal wave. These compressions and rarefactions caused the flame to oscillate at the same frequency of the speaker (10Hz).
16. [2 marks] (VCE 2003 Q4)
The distance has been doubled from 10m to 20m. A doubling of the distance means that the intensity has decreased to of the original. This means that the sound level L0 has dropped by 6dB
\ D (ANS)
17. [2 marks] (VCE 2004 Q2)
If the horizontal scale is 1 division = 100ms, then the number of divisions between cycles is 6.
The period is 5 ´ 100ms = 5 ´ 10-4s.
Frequency =
=
= 2000 Hz
= 2 kHz (ANS)
18. [2 marks] (VCE 2004 Q3)
If the distance is doubled the intensity is quartered.
\ Inew =
= 25mW m-2 (ANS)
19. [2 marks] (VCE 2004 Q4)
Each time the intensity is halved, the sound level drops by 3dB.
\ ¼ means ½ ´ ½ = -6dB (ANS)
20. [2 marks] (VCE 2004 Q9)
Chris (ANS)
This is the standard definition.
21. [2 marks] (VCE 2004 Sample Q5)
\B (ANS)
22. [2 marks] (VCE 2004 Sample Q6)
Sound is a longitudinal wave. The speaker created pressure variations which travelled away from the speaker as a longitudinal wave. There is no net displacement of the medium. The dust particle will only move back and forwards about its mean position.
23. [2 marks] (VCE 2004 Sample Q9)
This is found from using
L = 120 – 10log(5 ´ 10-3).
This should be clear from the other calculations.
\ 97dB (ANS)
24. [2 marks] (VCE 2004 Sample Q10)
If you sketch on the graph it will be around 300cm (ANS)
25. [2 marks] (VCE 2004 Sample Q11)
It is too difficult to get an estimate of the intersection from the graph, so you need to use the formula to come up with and answer.
Using the sound level at 10cm of 117dB. An increase of 3dB would give 120dB.
This means that the intensity has doubled so the distance has decreased by a factor of
10 ´ = 7.1cm (ANS)
26. [2 marks] (VCE 2004 Pilot Q2)
L (in dB) = 10 log
where L = 120 dB and I0 = 1 x 10-12
\ 120 = 10 log
\ 12 = log
\ 1012 =
\ I = 1
\ B (ANS)
It is wise to have some standard values of I and their equivalent values in dB on your formula sheet.
27. [2 marks] (VCE 2004 Pilot Q3)
Halving the sound intensity will result in a change of -3 dB.
\ 120 – 3 = 117
\ B (ANS)
28. [3 marks] (VCE 2005 Q1)
A sound wave is a [longitudinal] wave in which the air particles move [parallel to] the direction of propagation of the wave. The wave transmits [energy] from the source to the receiver.
29. [2 marks] (VCE 2005 Q10)
If the distance is double (200 m compared with 100 m) then the intensity must be
=.
Each change of means a -3dB change
\ a change of will be a –6dB change.
\ 6dB lower (ANS)
30. [2 marks] (VCE 2005 Q11)
Both balloons are the same distance from Balloon A. Therefore the sound intensity will be the same at both balloons B and C.
Use the intensity level formula
31. [2 marks] (VCE 2006 Q2)
Sound is a longitudinal wave. The speaker creates pressure variations which travel away from the speaker as a longitudinal wave. There is no net displacement of the medium. The dust particle will only move horizontally back and forwards about its mean position.
\ D (ANS)
32. [2 marks] (VCE 2006 Q3)
Using v = fl
\ l=
\ l =
\ l = 0.71 m (ANS)
33. [2 marks] (VCE 2006 Q4)
64 = 10 log
where I0 = 1 x 10-12
\ 6.4 = log
\ = 106.4
\ I = 106.4 ´ 1 ´ 10-12
\ I = 10-5.6
I = 2.5 ´ 10-6 W m-2 (ANS)
34. [2 marks] (VCE 2006 Q5)
If the distance is increased by a factor of 4
(8 m compared with 2 m) then the intensity must be
=.
Each change of means a -3dB change
\ a change of will be a –12dB change.
\ 64 - 12
\ 52dB
\ D (ANS)
35. [2 marks] (VCE 2008 Q2)
Sound is a longitudinal wave, so the movement of the particles is in the same direction as the motion of the wave. The dust particle is 10 cm in front of the speaker so it will move toward and away from the speaker but average the same distance from the speaker.
\ B (ANS)
36. [2 marks] (VCE 2008 Q3)
This is a straight forward application of the wave equation:
\ B (ANS)
37. [2 marks] (VCE 2008 Q4)
\ C (ANS)
38. [2 marks] (VCE 2008 Q5)
The intensity is inversely proportional to the radius squared, . So the ratio of the intensities is equal to the inverse ratio of the radiuses squared.
\ B (ANS)
39. [2 marks] (VCE 2008 Q6)
On the graph look at the 60dB on the y-axis, follow it across until you get to the 10,000 Hz point and see which phon line intersects the 2. The phon line that crosses it is the 40 phon. This means that a person would only perceive the 60 dB as 40 phon.
\ B (ANS)
40. [2 marks] (VCE 2008 Q7)
The phon is the perception of the intensity by humans and it changes with the frequency.
\ C (ANS)