Lacey asked:
The pendulum in an antique clock can be approximated by a thin rod with a coefficient of thermal expansion of 2.04e-5 (C°)e-1. At 22.5° C, the period of the pendulum is 1.00 s. If the temperature increases to 32.1° C, how many seconds per day will the clock gain or lose
Miller’s Antiques Handbook and Price Guide 2010-2011
The pendulum in an antique clock can be approximated by a thin rod with a coefficient of thermal expansion of 2.04e-5 (C°)e-1. At 22.5° C, the period of the pendulum is 1.00 s. If the temperature increases to 32.1° C, how many seconds per day will the clock gain or lose
Miller’s Antiques Handbook and Price Guide 2010-2011
Tags: Antique Clock, Pendulum, Thin Rod
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The pendulum in an antique clock can be approximated by a thin rod with a coefficient of thermal expansion of 2.04e-5 (C°)e-1. At 22.5° C, the period of the pendulum is 1.00 s. If the temperature increases to 32.1° C, how many seconds per day will the clock gain or lose
Period = 2 * π * (L/g)^0.5
1 = 2 * π * (L/9.8)^0.5
L = 0.2482369 m
∆ Length = Original length * coefficient of thermal expansion * ∆ Temperature
∆ Length = 0.2482369 * 2.04 * 10^-5 * (32.1 – 22.5) = 4.861471 * 10^-5
New length = 0.2482369 + 4.861471 * 10^-5 = 0.2482855 m
Period = 2 * π * (0.2482855/9.8)^0.5 = 1.000097886
Increase of period = 0.00009786 seconds
The period is longer, so the clock is moving slower.
Losing time
The period T = 2 π √ L/g.
If T1 is the period when the temperature is 22.5° C and
T2 is the period when the temperrature is 32.1° C
T2/T1 = √ (L2/L1)
L2 = L1 (1 + αθ) where θ = (32.1-22.5) = 9.6° C
T2/T1 = √ (1 + αθ) = √ (1 + 2.04e-5*9.6) = 1.0000979 s
T2 = 1.0000979 since T1 = 1s.
For every 1.0000979 s this clock will show only 1s
One day has 86400 s, In 86400 s the clock will show 86400/ 1.0000979= 86392 s
Thus the clock will loose ~9s in each day.
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Pendulum period in seconds
T ≈ 2π√(L/g)
or, rearranging:
g ≈ 4π²L/T²
L ≈ T²g/4π²
L is length of pendulum in meters
g is gravitational acceleration = 9.8 m/s²
L = T²g/4π² = 0.248237 meters
∆L/L = α∆T, α is linear thermal expansion coef
I had to puzzle over your number for α for a while as it it totally meaningless. I’ll just assume it is 2.04e-5/ºC.
∆L = Lα∆T = (0.248237)(2.04e-5)(32.1–22.5) = 0.0000486 m
new length is 0.24829
plugging that into the above formula for period, we get T = 1.0000981
so the difference is 0.0000981 sec
there are 3600×24 seconds per day, which gives you 8.5 sec error per day
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