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Chapter 8: Sound

Re: Chapter 8: Sound

Postby arexia » Fri Jul 28, 2017 9:41 pm

Hi,

For Q12 (in the online Q&A section), the answer says:
"Knowing that vY ∼ 1/(√ ρY), we get: New velocity (vX) = 1/√(2ρY) = 1/(√2) x 1/(√ρY) = 1/(√2) x vY Thus (vX)/(vY) = 1/√2"

But in the answer key, the ratio vX:vY is 1:√2.

Should it be 1:1/√2, or am I missing something?

Thanks
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Re: Chapter 8: Sound

Postby goldstanda3269 » Fri Jul 28, 2017 11:56 pm

arexia wrote:Hi,

For Q12 .... Thus (vX)/(vY) = 1/√2"

But in the answer key, the ratio vX:vY is 1:√2.

Should it be 1:1/√2, or am I missing something?


A fraction is a ratio, thus a fraction is a proportion,

(vX)/(vY) = 1/√2

can be translated thus: (vX) is to (vY) as 1 is to √2.

And so the answer is correct.

If you have the current edition of the GS book, see the GAMSAT Math section on proportions GM 1.6 and/or fractions GM 1.4.
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Re: Review Q6

Postby brad.willo2670 » Sun Apr 14, 2019 7:18 pm

Hi there,

Question 6: I understand the way in which the problem was solved, but I do not recall coming across learning the relationship between wave length and 'detection'? Which part of the text details this relationship?

Thanks in advance,
Brad
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Re: Review Q6

Postby goldstanda3269 » Mon Apr 15, 2019 2:05 pm

brad.willo2670 wrote:Question 6: I understand the way in which the problem was solved, but I do not recall coming across learning the relationship between wave length and 'detection'? Which part of the text details this relationship?


Good point. ACER often, but not always, would provide guidance for such a question. We have updated the question and somewhat modified the calculation based on recent trends. Enjoy!

NB: the next free webinar... www.gamsat-prep.com/GAMSAT-free-online-seminar/
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Re: Chapter 8: Sound

Postby Sandi » Sat Dec 07, 2019 5:31 pm

Hi there,

8.5.1 Doppler Effect Problem

In the book, the final answer is 17.9m/s but when I work out the calculation the answer is approximately around 142m/s because it is 86722/608 (according to my calculation). Am I doing something wrong?

Thanks,
Sandi
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Re: Chapter 8: Sound

Postby jeanne_gsgamsat » Sun Dec 08, 2019 9:33 am

Sandi wrote:Hi there,

8.5.1 Doppler Effect Problem

In the book, the final answer is 17.9m/s but when I work out the calculation the answer is approximately around 142m/s because it is 86722/608 (according to my calculation). Am I doing something wrong?


Yes, the problem is that you did not follow the proper order of operations. You can't ADD 277 directly to 331. You must first multiply -262 x 331 and then divide that number by 277 and then finally add that total to 331.

To be sure that you follow the proper order of operations, please consider going back to Chapter 1 GAMSAT Maths section 1.2.3.

Step by step:

vs = -262(331)/277 + 331
vs = -86722/277 + 331
vs = -313 + 331
vs = 18 m/s
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Re: Chapter 8: Sound

Postby Sandi » Tue Dec 10, 2019 10:45 pm

That makes sense! Thank you!
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Re: Chapter 8: Sound

Postby HunterEd » Thu Apr 09, 2020 1:03 am

6) Ultrasound is an important medical imaging tool. The wavelength used in ultrasound sets the limit to the smallest-sized objects that can be detected.

Consider a small kidney stone of 2.0 millimetres in size, approximately what minimal frequency of ultrasonic waves is necessary to detect this object? (Note: ultrasonic waves travel at approximately 1.5 x 103 m/s in the body)
A) 1.3 x 10-5 Hz
B) 130 kHz
C) 7.5 x 10-5 Hz
D) 750 kHz

Explanation: To be detected, according to the information provided, the wavelength of the wave must be of the order of the object’s size. For an object of 2.0 x 10-3 m in size (= 2 mm) to be detected (FYI: a small kidney or gallbladder stone), the wavelength should accordingly be: λ = 2.0 x 10-3 m.

f = v/λ (you don't need the equation; you can just note that Hz is the inverse of seconds and then use dimensional analysis, i.e., follow the units as done below)
f = (1.5 x 103 m/s) / (2 x 10-3 m)
f = (15 x 102 m/s) / (2 x 10-3 m)
f = 7.5 x 10^5 s-1 = 750 x 10^3 s-1
Recall that Hz = s-1, thus:
f = 750 x 103 Hz = 750 kHz

Correct me if i'm wrong but shouldn't this say, 750 x 10^4 s-1 instead of 750 x 10^3 s-1 ?

thank you
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Re: Chapter 8: Sound

Postby goldstanda3269 » Thu Apr 09, 2020 3:54 pm

HunterEd wrote:
f = 7.5 x 10^5 s-1 = 750 x 10^3 s-1
......

Correct me if i'm wrong but shouldn't this say, 750 x 10^4 s-1 instead of 750 x 10^3 s-1 ?


The number in the explanation is correct:

f = 7.5 x 10^5 s-1 = 750 x 10^3 s-1

Some students would remember it this way: If you add 2 decimal points to the factor (i.e. 7.5 to 750) then you must remove 2 from the power of ten (i.e. 10^5 to 10^3) in order to maintain equality.
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Re: Chapter 8: Sound

Postby HunterEd » Thu Apr 09, 2020 6:24 pm

oh wow, I must have had physics-brain yesterday. Sorry to waste your time!
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Re: Chapter 8: Sound

Postby goldstanda3269 » Fri Apr 10, 2020 2:23 pm

As long as you're studying/practicing, no time is wasted! Keep it up!
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Re: Chapter 8: Sound

Postby SeoM » Sat May 09, 2020 1:01 am

Hi, I'm really sorry but I am stuck on the explanation for the answer to question 10.

In the explanation, it says "10 log10 (a) = 20, or after simplification 10^log10(a) =a=10^2=100"

I don't understand the 'simplified' why 10^log10(a) =a . I know this is probably a basic log problem but I dont understand it and am hoping you could please shed some light on it for me please?
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Re: Chapter 8: Sound

Postby goldstanda3269 » Sun May 10, 2020 2:48 pm

SeoM wrote:Hi, I'm really sorry but I am stuck on the explanation for the answer to question 10.

In the explanation, it says "10 log10 (a) = 20, or after simplification 10^log10(a) =a=10^2=100"

I don't understand the 'simplified' why 10^log10(a) =a . I know this is probably a basic log problem but I dont understand it and am hoping you could please shed some light on it for me please?


Log rules are not so much about understanding but certainly need to be memorized. This particular rule is number 5 in the book ( GAMSAT Maths in the Gold Standard GAMSAT Book 2, GM 3.7 and 3.8 ) or you can see the rule online here:
https://www.gamsat-prep.com/gamsat-math ... hmicScales

You can also go to 13:47 in the General Chemistry video "New: Buffers" to see this specific log rule in action.

Basically, in order to isolate "a" in question 10, we need to get rid of the log base 10 and the only way to do this is to raise both sides to the power of 10 (this is rule 5 in GM 3.7.1).
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Re: Chapter 8: Sound

Postby NooraALUBAIDE » Thu Jan 19, 2023 11:25 pm

I would like to ask that as we have been informed before ( remember that tiny differences at the very top of log scales are massively more important than huge differences at the lower part of the scale)

so according to this information if I choose the range between 100 and 90 dB (20-30 Hz ) would it be wrong ? knowing that it's at the top of the scale too ? or the choice came in accordance with the given answers choices.
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Re: Chapter 8: Sound

Postby goldstanda3269 » Fri Jan 20, 2023 5:50 am

NooraALUBAIDE wrote:I would like to ask that as we have been informed before ( remember that tiny differences at the very top of log scales are massively more important than huge differences at the lower part of the scale)

so according to this information if I choose the range between 100 and 90 dB (20-30 Hz ) would it be wrong ? knowing that it's at the top of the scale too ? or the choice came in accordance with the given answers choices.


Which question are you referring to in GAMSAT Physics Chapter 8?
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