Question

A 0.870 m long brass pendulum experiences a 21.0°C temperature increase. (a) Calculate the original period (in s). s (b) What is the change (in s) in the period? s (c) Find the ratio (in h) of the change in period to the original period, and calculate how long a clock run by this pendulum would take to fall 1.00 s behind the correc

Answer 1

a)
Original period, To = 2 SQRT[Lo/g]
Where Lo is the original length of the pendulum
To = 2 x SQRT[0.870/9.807]
= 1.871 s

b)
Length of the pendulum after temperature increase, L = Lo[1 + T]
Where is the coefficient of thermal expansion of brass and T is the increase in temperature
= 19 x 10^-6 per degree
T = 21 ^oC

L = 0.87 [1 + (19 x 10^-6) x 21]
= 0.87034713 m

T = 2 x SQRT[0.87/9.81]
= 1.872 s

c)
Change in period, T - To = 1.872 - 1.871 = 0.001 s
Original period, To = 1.871 s

Ration of change in period to original period = 0.001/1.871
= 0.0005

Time needed to fall behind 1 s before correction = 1/0.005
= 1871 s