if a simple pendulum is driven clock gains 10 seconds per day, what fractional change in the pendulum length must be made for it to keep perfect time
2 answers
see this previous response for a slightly different error. https://www.jiskha.com/display.cgi?id=1510151111
the period of a simple pendulum is proportional to the square root of its length
P = k √L
the clock is gaining time , so the period is too short
seconds in a day ... 24 * 60 * 60 = 86400
... the error is one part in 8640 ... about a hundredth of a percent
P1 / √L1 = P2 / √L2 ... L2 = L1 (P2 / P1)^2 ... L2 = L1 {1 / [1 - (1 / 8640)]}^2
P = k √L
the clock is gaining time , so the period is too short
seconds in a day ... 24 * 60 * 60 = 86400
... the error is one part in 8640 ... about a hundredth of a percent
P1 / √L1 = P2 / √L2 ... L2 = L1 (P2 / P1)^2 ... L2 = L1 {1 / [1 - (1 / 8640)]}^2
1 answer