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16 March, 19:57

If the clock runs slow and loses 15 s per day, how should you adjust the length of the pendulum?

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  1. 16 March, 20:48
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    L = 1 m, ΔL = 0.0074 m

    Explanation:

    A clock is a simple pendulum with angular velocity

    w = √ g / L

    Angular velocity is related to frequency and period.

    w = 2π f = 2π / T

    We replace

    2π / T = √ g / L

    T = 2π √L / g

    We will use the value of g = 9.8 m / s², the initial length of the pendulum, in general it is 1 m (L = 1m)

    With this length the average time period is

    T = 2π √1 / 9.8

    T = 2.0 s

    They indicate that the error accumulated in a day is 15 s, let's use a rule of proportions to find the error is a swing

    t = 1 day (24h / 1day) (3600s / 1h) = 86400 s

    e = Δt = 15 (2/86400) = 3.5 104 s

    The time the clock measures is

    T ' = To - e

    T' = 2.0 - 0.00035

    T' = 1.99965 s

    Let's look for the length of the pendulum to challenge time (t ')

    L' = T'² g / 4π²

    L' = 1.99965 2 9.8 / 4π²

    L ' = 0.9926 m

    Therefore the amount that should adjust the length is

    ΔL = L - L'

    ΔL = 1.00 - 0.9926

    ΔL = 0.0074 m
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