Optimization Problem: PENDULUM SWINGING?

J

Joanne

New member
if t seconds is the time for one complete swing of a simple pendulum of length l feet, then 4pi^2 = gt where g=32.2. A clock having a pendulum of length 1 ft gains 5 minutes each day. Find the approximate amount by which the pendulum should be lengthened to correct the accuracy.
 
For a simple pendulum
T = 2π*sqrt(L/g)
dT/dL = π/sqrt(L*g)
dT = π/sqrt(L*g)*dL
∆T = π/sqrt(L*g)*∆L (approximately)
∆L = sqrt(L*g)*∆T/π

Now, what is ∆T? If the clock gains 5 minutes each day, the period is too short, meaning the length is too short. The error rate is
ε = +5/(24*60) = 3.4722e-3 [sec/sec]
∆T = ε*T
∆T = 2π*sqrt(L/g)*ε

Substituting
∆L = sqrt(L*g)*2π*sqrt(L/g)*ε/π
∆L = 2*L*ε
∆L = 2*1ft*3.4722e-3
∆L = .00694 ft = .0833 in
 
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