Question

A pendulum clock has a heavy bob supported on a very thin steel rod that is 1.00000 m long at 21.0 ∘C.

a)To 6 significant figures, what is the clock's period? Assume that g is 9.80 m/s2 exactly.

b)To 6 significant figures, what is the clock's period if the temperature increases by 13.0 ∘C? Assume that coefficient of linear expansion of steel is 1.20×10−5 K−1 .

c)The clock keeps perfect time at 21.0 ∘C. At 34.0 ∘C after how many hours will the clock be off by 1.00 s ?

Answer 1

a)
Time period, T = 2 SQRT[L/g]
Where L is the length of the pendulum and g is the acceleration due to gravity.
T = 2 x SQRT[1.00000/9.80]
= 2.00709 s

b)
Increase in length, L = LT
Where is the coefficient of expansion and T is the increase in temperature.
L = 1.00000 x 1.20 x 10^-5 x 13
= 0.00016 m

New length, L' = L + L
= 1.00000 + 0.00016
= 1.00016 m

New time period, T' = 2 x SQRT[L'/g]
= 2 x SQRT[1.00016/9.8]
= 2.00725 s

c)
T' - T = 2.00725 s -  2.00709 s
= 0.00016 s

Time exceed in 1 s = 0.00016 / 2.00725
= 0.000078 s

Time taken to reach 1 s = 1/ 0.000078
= 12822 s
= 12822 / 3600 h
= 3.56 h