A pendulum clock (close to the Earth’s surface) is placed in an elevator that mo

Question

A pendulum clock (close to the Earth’s surface) is placed in an elevator that moves upward with acceleration a for half of its ascent (distance d/2), and decelerates at -a during the second half of its ascent.
a) Is the elapsed time registered on the pendulum clock the same as what would register on an electronic watch? Explain your answer.
b) If your answer to (a) is “no”, derive an expression for the fractional difference in times: (Tpendulum - Twatch)/Twatch
c) expand your answer to second order in a/d.

Explanation / Answer

a. for distance d/2, acceleration = a

for distrance d/2, deceleration = a

now, Time period of pendulum clock = T = 2*pi*sqroot(l/g)

time recorded by digital watch while going up = t1

d/2 = 0.5*a*t1^2

t1 = sqroot(d/a)

time recorded by pendulum, t1'

t1' = t1*2*pi*sqroot(l/g)/2*pi*sqroot(l/(g+a))

t1' = t1*sqroot(1 + a/g)

time recorded by digital watch while going upo the other half = t2

v = a*t1 = a*sqroot(d/a) = sqroot(ad)

hence

sqroot(ad) = a*t2

t2 = sqroot(d/a)

t2' = t2*sqroot(1 - a/g)

hecne t1 + t2 - t1' - t2' = t1(1 - sqroot(1 + a/g)) + t2(1 - sqroot(1 - a/g))

dt = sqroot(d/a)(2 - sqroot(1 + g/g) - sqroot(1 - a/g))

since dt ! = 0

hencethe elapsed time on pendulum clocks is different than the digital clock

b. (Tpendulum - Twatch)/Twatch = -dT/2t1 = (sqroot(1 + a/g) + sqroot(1 - a/g) - 2)/2

c. expanding

(1 + a/2g - a^2/2g^2 + 1 - a/2g - a^2/2g^2 - 2)/2

-a^2/2g^2

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