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Thermal Expansion - Pendulum of a Grandfather Clock

  1. Sep 1, 2009 #1
    Thermal Expansion -- Pendulum of a Grandfather Clock

    1. The problem statement, all variables and given/known data
    The pendulum in a grandfather clock is made of brass (coefficient of linear expansion = 19 x 10^-6) and keeps perfect time at 17 degrees Celsius. How much time is gained or lost in a year if the clock is kept at 28 degrees Celsius? (Assume the frequency dependence on length for a simply pendulum applies.)


    2. Relevant equations
    T = 2pi * sqrt(L/g)
    delta L = alpha*L(naught)*delta T
    L = L(naught)*(1 + alpha*delta T)


    3. The attempt at a solution
    I really don't know where to even start with this. Once given a push in the right direction, I should be ok. But I just don't get the relationship.
     
  2. jcsd
  3. Sep 1, 2009 #2

    Hootenanny

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    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    Welcome to Physics Forums.

    HINT: What is the length of the pendulum if it keeps perfect time?
     
  4. Sep 1, 2009 #3
    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    I'm not sure. Wouldn't it depend on the gravity at a certain location? The instructor posted a worksheet that compares the T = 2pi * sqrt(L/g) initial to the final. But the comparisons he made are foreign to me. He has the fractional loss given by delta T/T(naught). Then he does a lot of substitutions and simplifications that I don't completely follow.
     
  5. Sep 1, 2009 #4

    Hootenanny

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    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    Okay, let's start by finding the ratio T/T0 then.
     
  6. Sep 1, 2009 #5
    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    Ok, so I have T0 = 2pi * sqrt(L0/g) at 17o C and T = 2pi * sqrt(L/g) at 28o Celsius. T is greater than T0 because L is greater than L0. I understand this part. But then they have the clock is losing time by an amount of delta T = T-T0. Where does that come from?
     
  7. Sep 1, 2009 #6

    Hootenanny

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    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    Good. Now, the next step is to find the fractional change in period, as your instructor did. So can you write down

    [tex]\frac{T}{T_0} = \ldots[/tex]
     
  8. Sep 1, 2009 #7
    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    T/T0 = 2pi * sqrt(L/g) / 2pi * sqrt(L0/g)

    = sqrt(L/g) / sqrt(L0/g)

    = sqrt(L) / sqrt(L0)

    = ???

    Now i'm lost again.
     
  9. Sep 2, 2009 #8

    rl.bhat

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    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    [tex]\frac{T}{To}[/tex] = sqrt( Lo + Lo*α*t)/sqrt(Lo)
    Length Lo of the grandfather clock is 1 m and its period is 2 s.
    Since [tex]\alpha[/tex] is very small, by using binomial expansion, the equation can be rewritten as

    [tex]\frac{T}{To}[/tex] = 1 + 1/2*α*t, where t is the temperature difference and α is the coefficient of linear expansion.

    [tex]\frac{T}{To}[/tex] - 1 = 1/2*α*t

    [tex]\frac{T - To}{To}[/tex] = 1/2*α*t

    (T - To)/To = 1/2*α*t = the chage in then period per second.
    (T - To) is the change in one oscillation, i.e. in 2 seconds.
    ( T - To )/To* number of seconds in one year gives you the required result.
     
    Last edited: Sep 2, 2009
  10. Sep 2, 2009 #9
    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    Where do you get the information that T = sqrt(L0 + L0*a*t) ?
     
  11. Sep 2, 2009 #10

    Hootenanny

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    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    Okay, I'll walk through it slowly. We start from your first line,

    T/T0 = 2pi * sqrt(L/g) / 2pi * sqrt(L0/g)

    [tex]\frac{T}{T_0} = \frac{2\pi\sqrt{L/g}}{2\pi\sqrt{L_0/g}}[/tex]

    Notice that the factors of 2 pi cancel and we can combine the square root.

    [tex]\frac{T}{T_0} = \sqrt{\frac{L}{g}\frac{g}{L_0}}[/tex]

    The factors of g cancel yielding

    [tex]\frac{T}{T_0} = \sqrt{\frac{L}{L_0}}[/tex]

    You with me so far?

    The next step is to insert the expression for L in terms of L0, T and a into the expression. Do you follow?
     
  12. Sep 2, 2009 #11
    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    Yes, I got that far. So now we use L = L0(1 + alpha * delta T) ?


    sqrt(L0(1 + alpha * delta T)/L0) = sqrt(1 + alpha * delta T)

    alpha = 19 x 10-6
    delta T = 28 - 17 = 11

    sqrt(1 + (19 x 10-6 * 11)) = sqrt(1 + 2.09 x 10-4)

    sqrt(1.000209) = 1.0001044.....

    So is this number right? If so, now what?
     
  13. Sep 2, 2009 #12

    Hootenanny

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    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    Try re-reading rl.bhat's post, paying particular attention to the binomial expansion part.
     
  14. Sep 2, 2009 #13
    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    OK, I got the part where sqrt(L0(1 + alpha * delta T)/L0) would equal sqrt((L0 + L0* alpha * delta T)/T0).

    What I don't understand is where the 1/2 comes from. Does it have something to do with the original length being 1m and the original period equalling 2s?
     
  15. Sep 2, 2009 #14

    rl.bhat

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    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    If x is very small ( 1 + x )^1/2 can be written as 1 + 1/2*x, by neglecting the higher powers of x in the binomial expansion.
     
  16. Sep 3, 2009 #15
    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    I THINK I GOT IT!!!

    Ok,
    Original Period = 2pi*sqrt(L0/g) at 17 Celsius
    New Period = 2pi*sqrt(L/g) at 28 Celsius

    delta T/T0 = T-T0/T0 = (T/T0)-1

    [2pi*sqrt(L0/g) / 2pi*sqrt(L/g)] -1 = sqrt(L/L0) -1 = sqrt(L0(1+alpha*delta temperature)/L0) -1

    sqrt(1 + alpha*delta temperature) - 1 = sqrt(1 + 19*10^-6*11) - 1 = 1.044*10^-4

    1.044*10^-4 * 31536000s = 3295.339s = 54.9 minutes!

    My only question now is: Where did I pull "delta T/T0" from? I saw it on another website and used it to get the right answer, but I don't understand how it fits. I know that since its getting heated, the pendulum will expand, making L > L0 and thus T > T0. Am I missing something elementary? I feel like i'm looking over the obvious.

    Thanks for everyone help.
     
  17. Sep 3, 2009 #16

    rl.bhat

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    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    sqrt(1 + 19*10^-6*11) - 1 = 1.044*10^-4

    Check this calculation.
     
  18. Sep 3, 2009 #17
    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    That caculation checks out. Let me try to type it better.

    {sqrt[1+(19*10^-6)*11]} - 1 = sqrt(1.000209) - 1 = 0.0001044 = 1.044*10^-4
     
  19. Sep 3, 2009 #18

    rl.bhat

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    Re: Thermal Expansion -- Pendulum of a Grandfather Clock

    OK. That is correct.
     
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