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Tuning Fork Problem, Please help?

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Sally and Tim are using a closed air column to determine the speed of sound in the classroom. They hear a resonant length 51 cm from the top of the tube then another one when the water is 68 cm from the top of the tube by vibrating a 1024 Hz tuning fork over the column. What is the speed of sound in the classroom?

    2. Explain how Mervin could tune a string to 512 Hz on his guitar using a 512 Hz tuning fork. [4]

    3. Normal breathing has an intensity of 10 dB. The music at an average rock concert is 120 dB. How many times louder is the rock concert?

    2. Relevant equations

    3. The attempt at a solution

    I don't know the first two, but is number 3) 10^10 dB?
    I need help on 1 and 2
  2. jcsd
  3. Jan 20, 2010 #2
    You are looking at a standing wave in number one. The distance between two adjacent nodes is [tex]\frac{\lambda}{2}[/tex]

    You are missing the relevant equation for decibels. [tex]\beta=10log_{10}(\frac{I}{I_0})[/tex] is relevant to number three. [tex]I_0[/tex] is a reference intensity, for the dB scale, [tex]I_0[/tex] is the hearing threshold. Intensity is roughly equivalent to loudness. How many times more or less is always asking for a unitless ratio.

    Always go to class without distractions, take notes, and ask questions.
  4. Jan 20, 2010 #3

    Andrew Mason

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    1) What does the distance between the resonant lengths represent in terms of a wavelength of sound? The answer to that will allow you to determine the wavelength, from which you can determine speed using the universal wave equation.

    2) Think 'beats'.

    3) Explain your answer. What is the meaning of 'loudness' in terms of power? What does a factor 120:10 represent in terms of power ratio between these two sounds?

  5. Jan 20, 2010 #4
    number 1 doesn't make sence, because i thought closed air column was wavelength of 1/4, 3/4, 5/4

    So .51 m / .25 = lambda 2.04
    and .68 / .75 = lambda .91

    And this is not consistant wave lengh to use the Universal wave lengh equation

    I am so confused

    2. Can use please expand on beats?

    3. I thought if breathing is 10 dB then concert is 120 db the loudness distance has difference of 10 dB and i thought there for the rock concert was 10^10 dB.
    Is this right?
  6. Jan 20, 2010 #5
    1. Yes, but it doesn't apply here. The distance between two nodes is half lambda. [tex]\frac{3\lambda}{4}-\frac{\lambda}{4}=\frac{\lambda}{2}, \frac{5\lambda}{4}-\frac{3\lambda}{4}=\frac{\lambda}{2}[/tex]...

    2. Go read sympathetic resonance on Wikipedia. The answer is there. Beats may be useful, I have not needed to tune a guitar using this or any other method.

    3. Is wrong. "Times louder" is asking for a ratio without units. The ratio you want is in the logs. But you are only off by an order of magnitude and a unit. Which means you did something almost right.
    Last edited: Jan 20, 2010
  7. Jan 20, 2010 #6

    Andrew Mason

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    Homework Helper

    You had it almost figured out. The difference between .68 m and .51 is the distance between consecutive anti-nodes (ie. the distance between a peak and a trough), which is ___ of a wavelength. So if .68-.51 = .17m = ___ wavelength, then one wavelength = ___ m. Since v = frequency x wavelength. v = _____.

    If the string and tuning fork have slightly different frequencies, there will be a pattern of constructive and destructive interference that will result in a pattern of alternating loud and soft intensity sound, called beats. The beat frequency is the difference between the frequencies of the two sounds. If the two sounds are at exactly the same frequency, the beats will disappear.

    How do you figure a difference of 10db? The difference is 110 db.

    A difference of 10db between two sounds means the ratio of the power of the sound waves is 10^?. A difference of 20 db is a power ratio of 10^?. A difference of 110 db is a power ratio of 10^___.

    Hint: Let P_0 be the power in the quietest audible sound.

    [tex]Loudness_{rock} = 120db = 10log_{10}\frac{P_{rock}}{P_0} = 10(log_{10}P_{rock} - log_{10}P_0)[/tex]

    [tex]Loudness_{breathing} = 10db = 10log_{10}\frac{P_{breathing}}{P_0} = 10(log_{10}P_{breathing} - log_{10}P_0)[/tex]

    Subtract the second from the first to get the expression for the difference in loudness (110 db) in terms of the power of the sounds.

  8. Jan 20, 2010 #7
    I don't understand this because we haven't learned the log equation is their a simpler way
  9. Jan 20, 2010 #8
    if the distance between consecutive anti-nodes is distance between a peak and a trough 1/4 of a wavelength. So if .68-.51 = .17m = 1/4__ wavelength, then one wavelength = ___ m. Since v = frequency x wavelength. v = _____.

    im sorry

    i don't know where i'm going with this
  10. Jan 20, 2010 #9
    Waite, maybe i got it figured out:
    .68=.51=.17m which is 1/2 wavelength than .17*2 = .34m
    =.34*1024 Hz
    =348.16 m/s

    is this right?
  11. Jan 20, 2010 #10
    Between Andrew and Wikipedia, you have the answer to two.
    You have one.
    Logrithum formulas:
    [tex]a^b=c\Leftrightarrow b=log_a(c)[/tex]
    Use these to find the ratio on the inside. You will need some but not all of them for three. But you may find them all useful at some point in the future. There is no other way that I am aware of.
  12. Jan 20, 2010 #11
    but we have never been taught this equation i am in grade 11 physics
  13. Jan 20, 2010 #12
    are you saying that 348.16 isn't right, i mean i have really tried to do these and its not working, can i please have the answer so i will know what i am doing wrong
  14. Jan 20, 2010 #13
    You got the correct answer for 1.
    Andrew and Wikipedia have your answer for 2.
    [tex]10^{\frac{\beta_2-\beta_1}{10}}=\frac{I_2}{I_1}[/tex] The formula YOU should have worked out from my first reply and my last reply is right here.
    Beta is the intensity of the sound in decibels. I is the intensity in W/m^2. I/I is the unitless ratio you need.
  15. Jan 20, 2010 #14
    I'm sorry for not understanding but thank you for helping me i really appreciate it.
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