## 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**## 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...

## 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**## 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.

## 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**## 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).

## 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**## 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.

## 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**## 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

## 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**## 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.

## 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**## 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.## 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.