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Thread: how would I do this.....

  1. #1
    Oct 2006

    Question how would I do this.....

    The time it takes for a pendulum to complete the one period is simplified by the formula: T=2 pi square root (L/g) Where g is the constant measuring 9.80655m/s^2 and L is the length of the pendulum. The length of the pendulums from a manufacture are normally distributed with a mean of 10cm and a standard deviation of 0.01 cm. Assuming the at a 10cm pendulum gives the correct time, what is the probability that a clock using one of theses pendulums will lose more than 1 minute a day?
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  2. #2
    Junior Member
    Jun 2007
    Cambridge, UK
    Well, I would use this general result on 'combining errors' in things that are normally distributed:

    If you have a variable Z normally distributed about a mean value \bar{Z} with variance \sigma_{Z}^2, and a function F(Z), the function will be approximately normally distributed about \bar{F} = F(\bar{Z})with a variance

    \sigma_A^2 = \left( \frac{\partial F}{\partial Z} \right)^2 \sigma_Z^2

    In this case, the variable is the length L with \bar{L} = 10cm, \sigma_L^2 = (0.01cm)^2.

    The function is the period T(L) given by:

    T = 2\pi \sqrt{\frac{L}{g}}

    so \frac{\partial T}{\partial L} = \pi \sqrt{\frac{1}{gL}}.

    You need to find the standard deviation \sigma_T where \sigma_T^2 = \left( \frac{\partial T}{\partial L} \right)^2 \sigma_L^2 = \pi^2 \frac{1}{gL} \sigma_L^2.

    Use L = \bar{L} in that expression.

    ...and then, using the properties of the normal distribution, you can find the probability of T being wrong by a certain amount (enough to make the clock slow down by a minute a day). There's quite a bit of calculating left for you to do, but hopefully that should help you on your way.

    - - -

    You may have a different way to do it... it really depends on what you've been taught.
    Last edited by Pterid; Jun 6th 2007 at 06:48 AM. Reason: cleanup
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