Ch 10.2 Quiz
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Ch 10.2 Quiz
1.) A car speeds away with 139.6 km/hrfrom a factory which emits a whistle sound at 1,318 Hz. What is the frequency registered by the driver of the car (see sheet 12,use 343 m/s for the speed of sound)? Indicate with a negative (positive) sign whether the frequency registered would be higher (lower) if the car moved toward the station
2.) You approach a wall while emitting sound at 326 Hz. The sound is reflected from the wall back at you and you register a reflected frequency of 551.5 Hz. What is your speed (see sheet 14,15,use 343 m/s for the speed of sound)?
3.) The length of a guitar string is 0.5817 m. What is the maximum wavelength of a standing wave on this string (see sheet 19) ? Indicate with a positive (negative) sign whether this is the only allowed wavelength (there are many other allowed wavelengths which are fractions of the wavelength you calculated).
4.) The traveling wave velocity on a string is 124.48 m/s and the longest wavelength of a standing wave on the string is 1.603 m. What is the frequency (see sheet 19) ? Indicate with a negative (positive) sign whether you calculated the minimum (maximum) frequency of a standing wave on this string.
2.) You approach a wall while emitting sound at 326 Hz. The sound is reflected from the wall back at you and you register a reflected frequency of 551.5 Hz. What is your speed (see sheet 14,15,use 343 m/s for the speed of sound)?
3.) The length of a guitar string is 0.5817 m. What is the maximum wavelength of a standing wave on this string (see sheet 19) ? Indicate with a positive (negative) sign whether this is the only allowed wavelength (there are many other allowed wavelengths which are fractions of the wavelength you calculated).
4.) The traveling wave velocity on a string is 124.48 m/s and the longest wavelength of a standing wave on the string is 1.603 m. What is the frequency (see sheet 19) ? Indicate with a negative (positive) sign whether you calculated the minimum (maximum) frequency of a standing wave on this string.
Kathleen Guest
Question 1
Question 1:
Answer is NEGATIVE, because the closer you move to the station, the higher the frequency
You use the equation f' = ((vv_o)/(v)) * f
There is a minus sign because the observer (v_o, the car) is receding and moving
Now with numbers
A car speeds away with 217.7 km/hrfrom a factory which emits a whistle sound at 1,401 Hz. What is the frequency registered by the driver of the car (see sheet 12,use 343 m/s for the speed of sound)?
FIRST, convert the km/hr to m/s by multiplying by 1000 m/km / 3600 s/hr
217.7 x (1000/3600) = 60.4722
f' = 343  60.4722 / 343 = .82369 * 1401 = 1153.998
Answer is NEGATIVE, because the closer you move to the station, the higher the frequency
You use the equation f' = ((vv_o)/(v)) * f
There is a minus sign because the observer (v_o, the car) is receding and moving
Now with numbers
A car speeds away with 217.7 km/hrfrom a factory which emits a whistle sound at 1,401 Hz. What is the frequency registered by the driver of the car (see sheet 12,use 343 m/s for the speed of sound)?
FIRST, convert the km/hr to m/s by multiplying by 1000 m/km / 3600 s/hr
217.7 x (1000/3600) = 60.4722
f' = 343  60.4722 / 343 = .82369 * 1401 = 1153.998
Kathleen Guest
Question 3
wavelenght/2 = L so rearrange it and you get 0.xxxm*2 (0.xxxx is whatever your meters are)
ruipatel Guest
Re: Ch 10.2 Quiz
Anyone figure out how to do 2 or 4 yet?
I've been trying to figure out 2 for some time now but can't seem to get it right.
I figured the equation would be something like f' = (v) / (v+vs) * f
Where vs is the source (you) and there is a + because you are approaching
I thought it might be
frequency registered = (343)/(343+vs) * frequency emitted
and then solve for vs but that doesn't seem to be right
Maybe we can all combine our methods on how we thought to do it to figure it out?
I've been trying to figure out 2 for some time now but can't seem to get it right.
I figured the equation would be something like f' = (v) / (v+vs) * f
Where vs is the source (you) and there is a + because you are approaching
I thought it might be
frequency registered = (343)/(343+vs) * frequency emitted
and then solve for vs but that doesn't seem to be right
Maybe we can all combine our methods on how we thought to do it to figure it out?
Kathleen Guest
Question 2
Okay so for #2, I'm not really sure why but you do
(frequency registered by you  frequency emitted) / (frequency registered by you + frequency emitted) = x
And then do X * frequency of sounds which is 343 m/s
You approach a wall while emitting sound at 344.3 Hz. The sound is reflected from the wall back at you and you register a reflected frequency of 541.1 Hz. What is your speed (use 343 m/s for speed of sound)?
So for example, it would be 541.1  344.2 / 541.1 + 344.2 = .2222 * 343 = 76.2394 m/s
(frequency registered by you  frequency emitted) / (frequency registered by you + frequency emitted) = x
And then do X * frequency of sounds which is 343 m/s
You approach a wall while emitting sound at 344.3 Hz. The sound is reflected from the wall back at you and you register a reflected frequency of 541.1 Hz. What is your speed (use 343 m/s for speed of sound)?
So for example, it would be 541.1  344.2 / 541.1 + 344.2 = .2222 * 343 = 76.2394 m/s
Kathleen Guest
Re: Ch 10.2 Quiz
yeah i just realized it doesn't work for the maple ta but it gave me the right answer 2 times on the practice quiz
Guest17 Guest
Re: Ch 10.2 Quiz
actually it does work i just typed in the number wrong the first time i did it on maple ta so that should work
Guest17 Guest
Re: Ch 10.2 Quiz
here is the problem given worked out maybe this will help
1.603*2 = 3.206 =L
v/(2L)
124.48/ (2*3.206)= 19.41
19.41*4=77.65
1.603*2 = 3.206 =L
v/(2L)
124.48/ (2*3.206)= 19.41
19.41*4=77.65
Guest17 Guest
Re: Ch 10.2 Quiz
yes, thank you. see i multiplied the length by 2 for L, but then i didn't multiply it by two again under the division sign. Quick question though, why is it multipled by four though? Like, how did you know where to get that from?
confused Guest
Re: Ch 10.2 Quiz
I have no idea I played around with it and I got the right answer when i multiplied it by four. the sheet says something about n being the number of wavelengths so i just started multiplying by each number until i got the answer because i wasn't sure how to find out how many wavelengths.
guest17 Guest
Q#3
#3 is simple,
just take the meters which is .5817 times it with 2 and that would be your answer (nagative answer)
.5817*2= ()1.1634
hope this helps. so sleep tight now:)
just take the meters which is .5817 times it with 2 and that would be your answer (nagative answer)
.5817*2= ()1.1634
hope this helps. so sleep tight now:)
Elephant Guest
Re: Ch 10.2 Quiz
I also didn't really understand that explanation for number two, so I did it this way...
f''=[(V+Vm)/(VVm)]f
The wall is both moving towards you as you are towards it. You can also convince yourself that because f'' is larger than f, you must add Vm on top and subtract Vm on the bottom so that the fraction will be greater than one. Then you just do some algebra and multiply it out. Both ways work...
f''=[(V+Vm)/(VVm)]f
The wall is both moving towards you as you are towards it. You can also convince yourself that because f'' is larger than f, you must add Vm on top and subtract Vm on the bottom so that the fraction will be greater than one. Then you just do some algebra and multiply it out. Both ways work...
Guesto Guest
#3
The maximum wave length is already given to you in m. So it is simply v/length.
Do not multiply the length by anything, there is no need to.
answer in neg.
Do not multiply the length by anything, there is no need to.
answer in neg.
super Mo Guest
Question 4
i also tried number 4 a couple of times and it doesnt work for me either. I really don't understand why you multiply the fundamental by 4.
heidy Guest
Question 4
Hey i just figured it out the question already states the longest wavelength and all you need to do is divide the velocity by the longest wavelength ( which is given) so it is simply: f=v/L
Heidy Guest
Question #2
Question 2:
You should have two numbers, one ~300, the other ~500. This example uses 303.1 Hz and 557.7 Hz. 343 m/s is the accepted speed.
x = (557.7303.1)/(557.7 + 303.1) = 0.295771
Then,
v = xV = 0.295771 x 343 = 101.4495 m s^1
You should have two numbers, one ~300, the other ~500. This example uses 303.1 Hz and 557.7 Hz. 343 m/s is the accepted speed.
x = (557.7303.1)/(557.7 + 303.1) = 0.295771
Then,
v = xV = 0.295771 x 343 = 101.4495 m s^1
DJ Guest
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