3 ms·
The equation holds in imperial units as well. The length of the 2 second pendulum needs to be in feet AND the value of g in ft/sec2.
by phkahler 2y ago
The equation holds in imperial units as well. The length of the 2 second pendulum needs to be in feet AND the value of g in ft/sec2.
- SAI_Peregrinus 2y agoπ^2 ≈ 32 to you?
- koolala 2y agoReplace s in your calculation with imperial s instead of metric s and it isn't imperial feet per metric seconds.
- SAI_Peregrinus 2y agoImperial seconds were very, very close to metric seconds.
- phkahler 2y agoSolving the equation for pi we get: PI = sqrt(g/L) g = 9.81. L=1 or g = 32.174. L=3.174 Either way works to approximately pi. There is a particular length where it works out exactly to pi which is about 3.2 feet, or about 1 meter. My point was that equations like that remain true regardless of units. The reason pi squared is approximately g is that the L required for a pendulum of 2 seconds period is approximately 1 meter.