Determine the change in gravity g in order to correct a clock with a seconds pendulum which is losing 10 seconds/day at a place where ݃ = 32 ft/sec2
Answers
Given : a clock with a seconds pendulum is losing 10 seconds/day where g = 32 ft/sec²
To find : Determine the change in gravity g in order to correct clock
Solution:
T = 2π √(L / g)
T = time Period
Pendulum is loosing 10 Secs per day
Hence Time lost = 10 Sec in 24 Hrs ( 86400 secs)
=> Time taken by pendulum = 86400 + 10 = 86410 Sec
86410 / 86400 = 2π √(L / 32) / 2π √(L / g)
=> 86410 / 86400 = √g/32
=> g = 32.0037
Change in gravity to correct time period = 32.0037 - 32
= 0.0037
Change in % = (0.0037/32)* 100 = 0.0116 %
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Answer:
One day is 86400 seconds. If your clock is losing five seconds per day, then it’s only counting out 86395 seconds in that span. In other words, your clock is slow by a factor of:
[math]\frac{86395}{86400} = 0.99994212963[/math]
The period of a pendulum is given by the equation [math]T = 2\pi \sqrt{\frac{l}{g}}[/math]
I don’t know how long your original pendulum is, so we’ll solve this through a ratio:
[math]\frac{T}{T_o} = 0.99994212963 = \frac{2\pi \sqrt{\frac{l}{g}}}{2\pi \sqrt{\frac{l_o}{g}}}[/math]
[math]0.99994212963 = \sqrt{\frac{l}{l_o}}[/math]
[math]l_o = \frac{l}{\sqrt{0.99994212963}}[/math]
This is a really, really small effect. You’re better off adjusting the value of [math]g[/math] — find a place where gravity is just a touch stronger.