phy 105

Derive the equation T = 2pi(l/g)^1/2.
That's period of motion.

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  1. hang mass m (will not matter) from string of length L

    deflect angle A from up and down
    Tension T in string
    m g = T cos A
    horizontal restoring force = -T sin A
    so
    F =-( m g/cosA) sin A = - m g tan A
    if A is small
    F = - m g A

    distance x from centered
    x = L sin A
    for small angle
    x = L A
    if x = s sin (2 pi t/T)
    v = s ( 2 pi/T) cos (2 pi t/T)
    a = - s (2 pi/T)^2 sin (2 pi t/T)
    or
    a = - (2 pi /T)^2 x

    now F = m a
    -m g A = -m(2 pi/T)^2 L A

    (2 pi/T)^2 = g/L

    2 pi/T = (g/L)^.5

    T = 2 pi (L/g)^.5

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