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Simple pendulum, and inertial and gravitational mass

  1. Mar 8, 2006 #1
    A simple pendulum has a length L. The inertial and gravitational masses of the bob are m1 and m2 respectively. Then the time period of the simple pendulum is given by
    T = 2(pie)[m1L/m2g]^(1/2) {Read as 2 pie root m one L by m two g)
    My question is that the ratio of inertial mass to gravitational mass is one, so why does m1 and m2 appear in the expression? Also, why do we put m1 in the numerator and m2 in the denominator? What I mean is that why couldn’t it have been in the following way,
    T = 2(pie)[m2L/m1g]^(1/2)
    Here the g is acceleration due to gravity and pie = 3.14 and the symbol ^ represents power.
  2. jcsd
  3. Mar 8, 2006 #2


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    Well, simply because m_{1} appears in the second law of Newtonian dynamics F=m_{1}a, while m_{2} appears in the universal attraction law F=-G m_{2}M/r^{2}...


    Eötvös proved experimentally the equality between the 2 and Einstein used this to formulate the Principle of Equivalence in GR.
  4. Mar 8, 2006 #3
    "Theoretical physics is a science locally isomorphic to mathematics"

    "mathematics is the language of theoretical physics :-)"
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