Calculate Fraction to Change Length of Pendulum for True Run

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To adjust the length of a pendulum that loses one minute every hour, the period must be shortened to allow for 60 swings in 60 seconds instead of 59. The correct relationship for the new period is T = t(1 - 1/60), indicating a fractional change in period rather than a direct subtraction. The discussion highlights the need to understand how the period relates to pendulum length using the equation t = 2π/√(g) * √(l). The user seeks clarification on this adjustment to proceed with the problem. Understanding this relationship is crucial for accurately calculating the required change in length for the pendulum to run true.
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Homework Statement


by what fraction must you change the length of the pendulum to make it run true if it loses one minute every hour

Homework Equations



t=2pi/sqrt(g) * sqrt(l)

The Attempt at a Solution


I know that I am losing 1 second every 60 seconds so I need to shorten the period by a certain amount so that it swings 60 times every 60 second instead of the 59 times that it currently swings.

so I want to say T=t-1/60 and then plug that into find L (T and L are final while t and l are initial).

but the correct answer is T=t(1-1/60), I don't know why this is or how we get it. I can do the rest of the problem if I can understand this one thing.

thanks,
 
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Does this work?

The frequency f(L) is 1/T(L) = [1/2pi]*[g/L]^.5 = 3540/3600 we get 3540 cycles per 3600 seconds with length L. But we want,

f(L-L/a) = [1/2pi]*[g/L-L/a]^.5 = 3600/3600

?
 
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