Assignment 10 - Sound
Sound Wave in Water
A sailor strikes the side of his ship just below the waterline. He hears the echo of the sound reflected from the ocean floor directly below 2.09 s later. How deep is the ocean at this point?
1.63×103 m
1.63×103 m
Discussion
NEW
the velocity in seawater
velocity of sound in seawater according to wiki is 1560 m/s
NEW
im confused on how to solve this problem i though i could just
multiply the velocity sound travels though water by the number
of seconds it that to hear the sound to determine the distance
but i keep getting the wrong answer.
If someone could help me i'd appreciate it
NEW
I was having problems with this one too. I was using the
same equation and continued to get it incorrect until a
friend helped me. Turns out you multiply the velocity of
sea water by the time given and divide by 2 for the correct
answer.
Ex: (1560m/s * 1.86s)/2 = 1451m
Hope it helps...
the velocity in seawater
velocity of sound in seawater according to wiki is 1560 m/s
NEW
im confused on how to solve this problem i though i could just
multiply the velocity sound travels though water by the number
of seconds it that to hear the sound to determine the distance
but i keep getting the wrong answer.
If someone could help me i'd appreciate it
NEW
I was having problems with this one too. I was using the
same equation and continued to get it incorrect until a
friend helped me. Turns out you multiply the velocity of
sea water by the time given and divide by 2 for the correct
answer.
Ex: (1560m/s * 1.86s)/2 = 1451m
Hope it helps...
Nervous Students
The noise level in an empty examination hall is 38 dB. When 100 physics students are writing an exam, the sounds of heavy breathing and pens traveling rapidly over paper cause the noise level to rise to 55 dB (not counting the occasional groans). Assuming that each student contributes an equal amount of noise power, find the noise level when half of the students have left the examination hall. Do not enter the unit.
5.21×10
5.21×10
Discussion
NEW
Anonymous 1
What equation do we use for this problem?
NEW
Does anyone know how to even start this? It says to leave
out the units, but we are solving for dB or...? How do we
go about finding the initial intensity?
NEW
Anonymous 1
I am still really confused on this problem. Any ideas?
NEW
Hmm Kiel T Sims
originally I thought that if you found the difference in dB
between the empty classroom and the full classroom and
divided that by 1/2. Then added that amount of dB to the
initial amount of dB then I would have the answer but it
isn't correct. Here are my calculations:
initial noise level: 38dB
noise level during exam: 60dB
noise added by students: 22dB
noise added by only 1/2 students: 11dB
final noise level: 49dB
Why did this not work?
NEW
Anonymous 4
can someone please explain how to do this problem? thanks alot
NEW
Anonymous 5
Can someone please explain how to do this problem?
NEW
"A LOT OF MONSTER TRUCKS"
Ooh, this one's a little tricky. If you're ever having
trouble guys, just remember to google it!
Anyway, what we want to do is solve for some I's. We're
given two different dB: one for the empty room, and one for
the room full of kids. For an example, let's say that the
empty room has an intensity of 40 Hz, and that the room
full of kids is, oh, 70 Hz. That means that the kids added
an intensity of 30 Hz to the room, and that 1/2 of those
kids will add 15 Hz to the room.
So what we want to do is solve for I when the room is
empty, I for when the room is full, and then find I when
the room is half empty.
40=10log(I/1*10^-12)= I(empty)
70=10log(I/1*10^-12) I(full)
subtract the two to find out the intensity of the children.
then divide by two to get the answer you need!
(I(full)-I(empty))/2
...i think
NEW
Anonymous 7
Hey guys here is the final equation that you need to use....
to find the level with half of the students just plug your
numbers into this equation: 10log(1/2(10^(full/10) - 10^
(empty/10)))
i found the explanation on this pdf page (it's number 7)
http://www.gantless.com/capa/1110/capa13.pdf
NEW
Here is some help Anonymous 8
http://oak.cats.ohiou.edu/~tt106402/work/phys252/Physics%20Notes/ch15.pdf
#59
NEW
Patrick is on the right track..
You just have to convert that last exponential number back
into dB. So, that last number you have is your I, plug it
back into xdB=10log(I/1x10^-12) and that is your answer...
don't put in the units.
Anonymous 1
What equation do we use for this problem?
NEW
Does anyone know how to even start this? It says to leave
out the units, but we are solving for dB or...? How do we
go about finding the initial intensity?
NEW
Anonymous 1
I am still really confused on this problem. Any ideas?
NEW
Hmm Kiel T Sims
originally I thought that if you found the difference in dB
between the empty classroom and the full classroom and
divided that by 1/2. Then added that amount of dB to the
initial amount of dB then I would have the answer but it
isn't correct. Here are my calculations:
initial noise level: 38dB
noise level during exam: 60dB
noise added by students: 22dB
noise added by only 1/2 students: 11dB
final noise level: 49dB
Why did this not work?
NEW
Anonymous 4
can someone please explain how to do this problem? thanks alot
NEW
Anonymous 5
Can someone please explain how to do this problem?
NEW
"A LOT OF MONSTER TRUCKS"
Ooh, this one's a little tricky. If you're ever having
trouble guys, just remember to google it!
Anyway, what we want to do is solve for some I's. We're
given two different dB: one for the empty room, and one for
the room full of kids. For an example, let's say that the
empty room has an intensity of 40 Hz, and that the room
full of kids is, oh, 70 Hz. That means that the kids added
an intensity of 30 Hz to the room, and that 1/2 of those
kids will add 15 Hz to the room.
So what we want to do is solve for I when the room is
empty, I for when the room is full, and then find I when
the room is half empty.
40=10log(I/1*10^-12)= I(empty)
70=10log(I/1*10^-12) I(full)
subtract the two to find out the intensity of the children.
then divide by two to get the answer you need!
(I(full)-I(empty))/2
...i think
NEW
Anonymous 7
Hey guys here is the final equation that you need to use....
to find the level with half of the students just plug your
numbers into this equation: 10log(1/2(10^(full/10) - 10^
(empty/10)))
i found the explanation on this pdf page (it's number 7)
http://www.gantless.com/capa/1110/capa13.pdf
NEW
Here is some help Anonymous 8
http://oak.cats.ohiou.edu/~tt106402/work/phys252/Physics%20Notes/ch15.pdf
#59
NEW
Patrick is on the right track..
You just have to convert that last exponential number back
into dB. So, that last number you have is your I, plug it
back into xdB=10log(I/1x10^-12) and that is your answer...
don't put in the units.
Open End Organ Pipe
Determine the length of an open organ pipe that emits middle C (262 Hz) when the temperature is 22.5°C.
0.657 m
0.657 m
Discussion
NEW
Anonymous 2
To solve:
Find velocity using: v=343+0.6(T-20deg)
(For T plug in temperature given in problem)
Then solve for Length using: L=v/2*f
(For v plug in velocity you solved for in previous
equation, f is the frequency given in problem).
Anonymous 2
To solve:
Find velocity using: v=343+0.6(T-20deg)
(For T plug in temperature given in problem)
Then solve for Length using: L=v/2*f
(For v plug in velocity you solved for in previous
equation, f is the frequency given in problem).
Tuning a Piano
A piano tuner hears one beat every 1.80 s when trying to adjust two strings, one of which is sounding 428 Hz, so that they sound the same tone. How far off in frequency is the other string?
0.556 Hz
0.556 Hz
Discussion
NEW
Anonymous 1
Any ideas on starting off this problem?
NEW
The formula is Fbeat=f1-f2
f1 is given but how do u convert 2.14 seconds into Hz to
find fbeat?
NEW
Re: Anonymous 3
guys its the # of beats over the # of seconds given.
NEW
Anonymous 4
can someone please explain how to do this problem? Thanks!
NEW
So my numbers are 459 Hz and 1 beat every 2.14 seconds. I
did 1/2.14 for the second frequency and got 0.467 Hz. I have
no idea what to do next and I only have one try left.
NEW
"A LOT OF MONSTER TRUCKS"
For some reason it's 1/number of seconds between each beat. I
had 2.2 as the seconds per beat, so 1/2.2=0.45 Hz. Units are
in Hz, and try to figure this out because I honestly have no
clue why that worked.
Anonymous 1
Any ideas on starting off this problem?
NEW
The formula is Fbeat=f1-f2
f1 is given but how do u convert 2.14 seconds into Hz to
find fbeat?
NEW
Re: Anonymous 3
guys its the # of beats over the # of seconds given.
NEW
Anonymous 4
can someone please explain how to do this problem? Thanks!
NEW
So my numbers are 459 Hz and 1 beat every 2.14 seconds. I
did 1/2.14 for the second frequency and got 0.467 Hz. I have
no idea what to do next and I only have one try left.
NEW
"A LOT OF MONSTER TRUCKS"
For some reason it's 1/number of seconds between each beat. I
had 2.2 as the seconds per beat, so 1/2.2=0.45 Hz. Units are
in Hz, and try to figure this out because I honestly have no
clue why that worked.
Beat Frequency
Two automobiles are equipped with the same single-frequency horn. When one is at rest and the other is moving toward an observer at 15.7 m/s, a beat frequency of 5.54 Hz is heard. What is the frequency the horns emit? Assume T = 20°C .
115 Hz
115 Hz
Discussion
NEW
HELP!
for some reason im gettin this wrong...im doing
f beat=f source(1/1- (v source/v sound))
and im solving for f source but im getting 5.15 and my
answer isnt working, help please!
NEW
i though i was doing this correctly but i also keep getting
the wrong answer
I took:
Pg.424
Fsource=
................................beat
frequency............................
_________________________________________ (Divided by)
(1/1-(velocity/343) -(1/1+(velocity/343))
but this is the incorrect answer, any suggestion would be
appreciated
HELP!
for some reason im gettin this wrong...im doing
f beat=f source(1/1- (v source/v sound))
and im solving for f source but im getting 5.15 and my
answer isnt working, help please!
NEW
i though i was doing this correctly but i also keep getting
the wrong answer
I took:
Pg.424
Fsource=
................................beat
frequency............................
_________________________________________ (Divided by)
(1/1-(velocity/343) -(1/1+(velocity/343))
but this is the incorrect answer, any suggestion would be
appreciated
Bat and Moth
A bat flies toward a moth at speed 8.12 m/s while the moth is flying toward the bat at speed 4.90 m/s. The bat emits a sound wave of 50.9 kHz. What is the frequency of the wave detected by the bat after it reflects off the moth?
5.49×101 kHz
5.49×101 kHz
Discussion
NEW
im not sure what im doing wrong
Ive been trying to use
eq2 For moving source and stationary observer
f"=f'/(1+ (vs/v))
eq1 For moving observer and stationary source
f'=f(1- (vo/v))
but i keep getting the wrong answer: 49.8kHz
here are my numbers
bat emits : 51800Hz
bat speed: 7.94m/s
moth speed: 5.02m/s
i used eq 1 and plugged in:
f'= 51800 (1 - (7.94/334))
f'=50568
then i took eq 2:
f"= 50568/ (1+ (5.02/334))
f"=49.8kHz
where did i go wrong?
thanks for the help
-Fahim
NEW
Re: im not sure what im doing wrong
I'm still figuring it out myself but I know that you need to
take into account that both there is no stationary source
since the bat and the both are both moving toward each other.
im not sure what im doing wrong
Ive been trying to use
eq2 For moving source and stationary observer
f"=f'/(1+ (vs/v))
eq1 For moving observer and stationary source
f'=f(1- (vo/v))
but i keep getting the wrong answer: 49.8kHz
here are my numbers
bat emits : 51800Hz
bat speed: 7.94m/s
moth speed: 5.02m/s
i used eq 1 and plugged in:
f'= 51800 (1 - (7.94/334))
f'=50568
then i took eq 2:
f"= 50568/ (1+ (5.02/334))
f"=49.8kHz
where did i go wrong?
thanks for the help
-Fahim
NEW
Re: im not sure what im doing wrong
I'm still figuring it out myself but I know that you need to
take into account that both there is no stationary source
since the bat and the both are both moving toward each other.
Speed Detector
A radar device emits microwaves with a frequency of 3.86E+9 Hz. When the waves are reflected from a van moving directly away from the emitter, the beat frequency between the source wave and the reflected wave is 605 beats per second. What is the speed of the van? (Note: microwaves, like all forms of electromagnetic radiation, propagate at the speed of light c = 3.00×108 m/s.)
8.46×101 km/hr
8.46×101 km/hr
Conceptual - Doppler
John is listening to a horn. He knows the frequency of the horn is 300 when both he and the horn are at rest. If he hears a pitch of 270 , there are clearly several possibilities. For each of the following statements, indicate whether it is correct or incorrect.
Correct: John is moving away from the horn at rest.
Incorrect: John and the horn are both moving in the same direction but John is behind and moving faster than the horn.
Correct: Both can be moving, in opposite directions.
Correct: Both can be moving and have the same speed.
Correct: Both can be moving and have different speeds.
Incorrect: The distance between John and the horn is decreasing with time.
Correct: John is moving away from the horn at rest.
Incorrect: John and the horn are both moving in the same direction but John is behind and moving faster than the horn.
Correct: Both can be moving, in opposite directions.
Correct: Both can be moving and have the same speed.
Correct: Both can be moving and have different speeds.
Incorrect: The distance between John and the horn is decreasing with time.
Discussion
NEW
Help Please Anonymous 1
Help please!
John is listening to a horn. He knows the frequency of the
horn is 300 $Hz$ when both he and the horn are at rest. If
he hears a pitch of 330 $Hz$, there are clearly several
possibilities.
The distance between John and the horn is increasing with time.
Both can be moving, in opposite directions.
John and the horn are both moving in the same direction but
John is behind and moving faster than the horn.
Both can be moving and have different speeds.
Both can be moving and have the same speed.
NEW
help with previous problems
for some odd reason everytime i answer a question the link
to post a thread goes away but i still wanted to help those
who were having a problem with "bat and moth" and "beat
frequency" is used:
http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PST4_3/PS
T4_3.htm
problem 86 and 54 respectively. hope it helps
NEW
Help Please Anonymous 3
Does anyone have numbers similar to mines, who can help me?
I would really appreciate it
John is listening to a horn. He knows the frequency of the
horn is 300 $Hz$ when both he and the horn are at rest. If
he hears a pitch of 270 $Hz$, there are clearly several
possibilities. For each of the following statements,
indicate whether it is correct or incorrect.
NEW
Help pls Anonymous 4
Cant figure this one out. These are the answers that i
thought were right but guess not. help pls
Conceptual - Doppler
John is listening to a horn. He knows the frequency of the
horn is 300 when both he and the horn are at rest. If he
hears a pitch of 330 , there are clearly several
possibilities. For each of the following statements,
indicate whether it is correct or incorrect.
INCORRECT: Both can be moving, in opposite directions.
INCORRECT: Both can be moving in the same direction at the
same speed.
INCORRECT: The distance between John and the horn is
increasing with time.
CORRECT: Both can be moving and have different speeds.
CORRECT: John is moving towards the horn at rest.
CORRECT: Both can be moving and have the same speed.
NEW
Answer!
A) John is moving towards the horn at rest.
B) Both can be moving and have the same speed.
C) Both can be moving and have different speeds.
D) Both can be moving, in opposite directions.
E) The distance between John and the horn is increasing with
time.
F) Both cannot be moving in the same direction.
ANSWER:
A)True
B)True
C)True
D)True
E)False
F)False
NEW
Correct answers to compare
John is listening to a horn. He knows the frequency of the
horn is 300 $Hz$ when both he and the horn are at rest. If
he hears a pitch of 270 $Hz$, there are clearly several
possibilities. For each of the following statements,
indicate whether it is correct or incorrect.
Correct: John and the horn are both moving in the same
direction but John is ahead and moving faster than the horn.
Correct: Both can be moving, in opposite directions.
Correct: John is moving away from the horn at rest.
Incorrect: The distance between John and the horn is
decreasing with time.
Correct: Both can be moving and have different speeds.
Correct: Both can be moving and have the same speed.
Help Please Anonymous 1
Help please!
John is listening to a horn. He knows the frequency of the
horn is 300 $Hz$ when both he and the horn are at rest. If
he hears a pitch of 330 $Hz$, there are clearly several
possibilities.
The distance between John and the horn is increasing with time.
Both can be moving, in opposite directions.
John and the horn are both moving in the same direction but
John is behind and moving faster than the horn.
Both can be moving and have different speeds.
Both can be moving and have the same speed.
NEW
help with previous problems
for some odd reason everytime i answer a question the link
to post a thread goes away but i still wanted to help those
who were having a problem with "bat and moth" and "beat
frequency" is used:
http://tuhsphysics.ttsd.k12.or.us/Tutorial/NewIBPS/PST4_3/PS
T4_3.htm
problem 86 and 54 respectively. hope it helps
NEW
Help Please Anonymous 3
Does anyone have numbers similar to mines, who can help me?
I would really appreciate it
John is listening to a horn. He knows the frequency of the
horn is 300 $Hz$ when both he and the horn are at rest. If
he hears a pitch of 270 $Hz$, there are clearly several
possibilities. For each of the following statements,
indicate whether it is correct or incorrect.
NEW
Help pls Anonymous 4
Cant figure this one out. These are the answers that i
thought were right but guess not. help pls
Conceptual - Doppler
John is listening to a horn. He knows the frequency of the
horn is 300 when both he and the horn are at rest. If he
hears a pitch of 330 , there are clearly several
possibilities. For each of the following statements,
indicate whether it is correct or incorrect.
INCORRECT: Both can be moving, in opposite directions.
INCORRECT: Both can be moving in the same direction at the
same speed.
INCORRECT: The distance between John and the horn is
increasing with time.
CORRECT: Both can be moving and have different speeds.
CORRECT: John is moving towards the horn at rest.
CORRECT: Both can be moving and have the same speed.
NEW
Answer!
A) John is moving towards the horn at rest.
B) Both can be moving and have the same speed.
C) Both can be moving and have different speeds.
D) Both can be moving, in opposite directions.
E) The distance between John and the horn is increasing with
time.
F) Both cannot be moving in the same direction.
ANSWER:
A)True
B)True
C)True
D)True
E)False
F)False
NEW
Correct answers to compare
John is listening to a horn. He knows the frequency of the
horn is 300 $Hz$ when both he and the horn are at rest. If
he hears a pitch of 270 $Hz$, there are clearly several
possibilities. For each of the following statements,
indicate whether it is correct or incorrect.
Correct: John and the horn are both moving in the same
direction but John is ahead and moving faster than the horn.
Correct: Both can be moving, in opposite directions.
Correct: John is moving away from the horn at rest.
Incorrect: The distance between John and the horn is
decreasing with time.
Correct: Both can be moving and have different speeds.
Correct: Both can be moving and have the same speed.