Question

The period, " T ", of oscillation in seconds of a pendulum clock is given by T = 2 * π * sqrt(l/g), where " g " is the acceleration due to gravity. The length of the pendulum, " l ", depends on the temperature, " t ", according to the formula l = l0 ( 1 + at ) where " l0 " is the length of the pendulum at temperature " t0 " and " a " is a constant which characterizes the clock. The clock is set to the correct period at the temperature " t0 ". How many seconds a day does the clock lose or gain whehttp://openn the temperature is t0 + Δt ? Show that this loss or gain is independent of l0

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Answer 1

The given answer for the first part of this problem is:

-432000 * α(alpha) * Delta * t / ( 1 + α(alpha) * t )^(3/2)

but I do not understand how to approach the problem in order to get to the answer.

Answer 2

T = 2*pi*sqrt(l/g)
or T= 2*pi*sqrt(l0(1+at)/g) (because l = l0(1+at))
at temperature t, the time period will be
T1= 2*pi*sqrt(l0(1+at)/g)
At t+del t, the Time Period will be T2 = 2*pi*sqrt(l0(1+a(t+del t))/g)
Change in time will be T2 -T1
Use binomial theorem to expand the term under square root

Answer 3

T1 = 2*pi*sqrt(l/g)
T2 = 2*pi*sqrt(l/g + l0*a*del t/g)
change in time = T2-T1