Question

A 0.810 m long brass pendulum experiences a 18.0°C temperature increase.

(a) Calculate the original period(s).

(b) What is the change in the period(s)?

(c) Find the ratio of the change in period to the original period, and calculate how long (h) a clock run by this pendulum would take to fall 1.00 s behind the correct time.

Answer 1

T₀=2πsqrt(L/g) =
=2πsqrt(0.81/9.8)=
=1.8064 s.
ΔL=α•L•ΔT=19•10⁻⁶•0.81•18
=2.77•10⁻⁴m
L₁=L+ ΔL =0.81+2.77•10⁻⁴=
=0.810277
T₁=2πsqrt(L₁/g)=
=2πsqrt(0.810277/9.8)=
=1.8067 s
T₁-T=1.8067-1.8064=0.0003 s.
(T₁-T)/T=0.0003/1.8064=1.66•10⁻⁴
(t/T₁)0.0003= 1 s
t= T₁/0.0003 = 1.8067/0.0003 =6022 s=1.67 h

Answer 2

L1 = 0.810 m.
Temp. Change = +18o C.
T.C. = 1.9*10^-5/C

a. T^2 = 4pi^2(L/g)
T^2 = 39.4(0.81/9.8) = 3.25653
T = 1.8046 s.

b. L2 = L1 + L1*(T.C.)*18
L2=0.81 + 0.81(1.9*10^-5)*18=0.81027702
m.

T^2 = 39.4(0.81027702/9.8)=3.25764
T = 1.8049 s.

Change in period = 1.8049-1.8046 = 0.00003 s.

c. 3*10^-5/1.8046 = 1.6624*10^-5