2008 Checkpoints Chapter 10 Sound Basics

2008 Checkpoints Chapter 10 Sound Basics

Question 432

You need to use the correct technique for answering multiple-choice questions. You are to train yourself to not look at the answers but to work your own out. Then see which of the answers agrees with you. You must always use this strategy, because there is usually one correct answer and 3 incorrect answers, the incorrect answers are specifically designed to 'look' correct. The other advantage is that when you find their answer that agrees with you, it is a confidence booster.

The wavelength is defined as the distance between two identical points on a wave-train (a series of waves).

It is usually measured from peak to peak, or trough to trough, only because these are easy to locate.

In a sound wave the compressions and rarefactions are shown, the wavelength is by definition, the distance between 2 rarefactions or 2 compressions.

Question 433

A wave is where the particles vibrate around a mean position, when they vibrate parallel to the direction of motion of the energy, they are called longitudinal, when they vibrate perpendicularly to the direction of motion of the energy then they are called transverse waves.

To gain 4 marks on an exam, the sentence use a diagram if you wish, must be read as, 'use a diagram if you want to get the full 4 marks'.

Transverse wave

the movement of the particles is in this direction

Longitudinal wave

The movement of particle is in this direction.

Question 434

0 5 10 15 distance from speaker (m)

pressure variation 3/4 period later.

If you consider the first peak in the black graph, i.e. the original line. It is a local maximum of pressure. 1/4 cycle later it is at normal atmospheric pressure, this would be represented by being on the horizontal axis.

A further 1/4 cycle, it would be a local minimum, represented by a trough.

A further 1/4 cycle it would be back to normal atmospheric pressure, i.e. on the horizontal axis.

If you consider the point at the origin, i.e. at d = 0. It will be at normal atmospheric pressure,

1/4 cycle later it will be a local minimum.

1/4 cycle later it will be back to normal.

1/4 cycle later it will be a local maximum.

So the orange graph represents the new pressure v distance graph.

Question 435

The wave equation is v = fl. We are given v = 330 m/s, the definition of the wavelength is the distance between two pressure maximums (also called compressions). So l = 5 m.

\ Using v = fl gives 330 = f × 5

\ f = 330/5

\ f= 66 Hz

Question 436

If the wave equation is v = fl. Then you need to use that the speed of sound is the same for all frequencies.

The longest wavelength will correspond to the shortest frequency, and the shorter wavelength will occur at the highest frequency.

For v = constant l µ

\ f20l20 = f20 000l20,000

\ =

= 20000/20

= 1000

Question 437

Your ear can hear sounds from a variety of sources at the same time. You are able to clearly distinguish the different sounds, eg you can hear someone speaking to you over the background noise of a stereo. The sound waves from both sources pass through each other without altering.

But you need to consider that the movement of the air molecules is a vector, so when the two waves are actually passing through each other their effects will add like vectors.

So the more complete answer is C, not A. In the exam you will always give the best answer. There will only be one answer, unless the question specifically says, 'one or more answers'.

Question 438

The voltage of the microphone represents the pressure fluctuations, so the distance between identical points on the graph is the wavelength of the sound wave.

It takes 4 ms (milliseconds) for the wave to complete a cycle. The frequency of the vibration is inversely related to the period. The longer the period the less cycles per second.

Using f = = = = 250 Hz

Question 439

If the wave equation is v = fl.

Then 345 = 250 × l

\ l = 345/250

= 1.38 m.

= 1.4 m

This answer is a little better written as 1.4 m, because from a sig. fig. perspective, you can only quote one figure in your answer. This would be very unwise, if you didn’t show your working out. Because an answer of 1, is not precise enough.

Question 440

If the wave equation is v = fl.

Then 336 = 512 × l

\ l = 336/512

= 0.65625 m.

= 0.66 m

Question 441

If the velocity is 336 m/s, then using v = d/t gives t = d/v

t = 3/336

t = 8.9 × 10-3 secs.

Again, if we consider sig. figs. then t = 9 × 10-3 secs.

Question 442

The speaker cone vibrates backwards and forwards. When it is moving forwards it compresses the air immediately in front of it, as it moves backwards it actually rarefies the air in front of it. It repeats this movement 512 times every second, because it is producing a sound with a frequency of 512 Hz.

These compressions and rarefactions travel away from the speaker, (the speaker continues to make new compressions and rarefactions) and eventually reach Freda's ear. The air doesn't actually move it is only the molecules vibrating about there normal positions. (The air molecules are also vibrating randomly all the time, the sound just imposes its own vibrations on top of these random motions)

The eardrum is sensitive to pressure variations and indirectly transmits these signals to the brain, which we interpret as sound.

Question 443

Sound travels as a wave. This wave is able to be reflected off hard walls. The bricks represent a hard flat surface, which will produce good reflection, which means that the reflected signal will be such that it is still able to be heard and recognized as sound.

If the wall was covered in an irregular surface, then the reflected signals become out of phase, we then get interference. If the reflecting wall was covered with something soft (like woollen carpet), then a lot of the signal is absorbed by the surface, and not much is reflected.

The sound wave will reflect in such a way that the angle of incidence is equal to the angle of reflection, this means that the person will hear an 'image' of the original source. This image will be in position 'Y', because it is the mirror image of the source.

Question 444

This is testing your understanding of the fact that sound will reflect of a hard surface, and that the angle of incidence will equal the angle of reflection.

This infers that the horizontal distance between A and B must equal the horizontal distance between B and C.

\ the distance AC = 340 × 2 = 680 m.

Question 445

340 m

480 m depth

Using Pythagoras, 4802 = 3402 + (depth)2

4802 - 3402 = (depth)2

depth = 338.8214869m

= 339 metres

Question 446

This question is not on the current course

The speed of sound in the higher temperature material would be greater, but the angle would stay the same. The speed of sound increases in materials as the temperature of the material increases, this is because as the temperature increases the average KE of the particles increase, which increases the speed of the energy travelling through the medium.

Question 447

The period = = = 0.625 = 0.63 secs

Question 448

Using v =

With the distance being 120m × 2, because the sound has to travel to the wall and back to Jennie before she claps again. So the sound travels 240 m each 0.625 seconds.

\ v = 240/0.625 = 384 m/s.

You need to make sure that this answer is reasonable; on the exam all answers should be reasonable answers. You should know that the speed of sound is around 340 m/s, and that it can vary with temperature. So this answer seems reasonable.

Question 449

If Jennie claps at half the frequency, there will be twice as long between claps. That means that after her first clap, the sound will travel to the wall, reflect back to her in 0.625 secs, she will then wait another 0.625 secs before she claps again, so she will hear the clap, then the 'echo' then the next clap, next 'echo', etc every 0.625 secs.

So she claps every 1.25 secs and hears an echo in between each clap.

Question 450

If she claps at a frequency of 3.2 Hz, she is clapping every 0.3125 secs.

The first clap will have travelled to the wall in this time. She claps again. By the time the second clap has reached the wall, the first clap has reflected and is back to where she is, so she hears the 'echo' after 2 claps. From here on she will hear the echo every time she claps, but it is the echo from 2 claps previous.

Question 451

Note that this question specifically states one or more answers. It means exactly that, there are either one or more answers, not necessarily more than one.

If the speed of sound was to rise, then it would take less time for the sound to travel to the wall and back. She would need to clap faster to keep the claps synchronised with the echo.

If she moved further away from the wall, it would increase the time it took for the sound to travel from her hands to the wall and back, this would mean that if the speed of sound increased, she would not have to clap any faster.

Question 452

Georgio's model is a better representation of a sound wave, because the slinky is demonstrating a longitudinal wave, the same as sound. The slinky is an excellent model to demonstrate sound and it clearly shows that the particles only vibrate around their mean position and that there isn't a net movement of material (be it air molecules or whatever). If sound did create a net movement of material, then you would expect all the slinky to end up and one end. This does not happen.

The vibration of the particles is parallel to the direction of travel of the propagation of the wave energy, this is the same as sound.

This model demonstrates the compressions as a bunching up of the slinky and the rarefactions as a stretching of the slinky.

Question 453

I think that they meant this question to read 'What is the sound level (in dB) 2.0 m from the passing train?

L (in dB) = 10 log

L = 10 log

\ L = 10 log 3.5 × 109

= 10 × 9.544068044

= 95 .4 dB

Question 454

This question is testing the sound intensity drops off as

So if the distance is changed from 2 metres to 10 metres then r is changed by a factor of 5.

This means that the intensity will go down by a factor of 52

52 = 25

So the new intensity will be the old intensity divided by 25

Inew = 3.5 × 10-3 / 25

= 1.4 × 10-4 W/m2

Question 455

If the sound intensity is increased by a factor of 4,

i.e. = 4

Then L = 10 log

L = 10 log 4

L = 10 × 0.602

L = 6.02 dB

L = 6 dB

This means that increasing the intensity by a factor of 4 increases the sound level by 6 dB. It follows that doubling the intensity will increase the sound level by 3 dB.

It also follows that decreasing the intensity by a factor of 4 decreases the sound level by 6 dB etc.

Question 456

L (in dB) = 10 log where I = 2 x 10-6 and I0 = 1 x 10-12

\ L = 10 log

= 10 log 2 x 106

= 10 × 6.3010

= 63 dB

Question 457

This question is testing the sound intensity drops off as

So if the distance is changed from 2 metres to 6 metres then r is changed by a factor of 3.

This means that the intensity will go down by a factor of 32

32 = 9

L (in dB) = 10 log

We are told that =

\ L = 10 log 9-1

= -10 × 0.9542

= -9.542

= -9.5 dB or a loss of 9.5dB.

Question 458

L (in dB) = 10 log

We are told that = 5

\ L = 10 log 5

= 10 × 0.6989

= 6.989

= 7 dB

Question 459

Tina is 35 metres from the source and she reads 5 x 10-4 W/m2. Jimi (obviously a thrill seeker) is a lot closer, and hence the sound will be a lot louder here.

The question wants you to use the inverse square law to find the new intensity.

It is basically a ratio question, and the intensity will increase by the factor (35/10)2.

The other way to find out is to work out what the intensity would be at 1 metre, using Tina's data, and then to use this new data to find out what it would be like at 10 m.