A pendulum clock has a pendulum shaft made of aluminum which has a coefficient of linear expansion of 24.0*10^-6K^-1. At 22 degrees celsius the period of the swing of the pendulum is 1.00seconds. If the temperature is reduced to 5 degrees celcius, how much time does the clock gain or lose in 1 day?

Question

what is its initial length?

Answer 1

Moment of inertia respectively the axis
that passes through the end of the pendulum is
I = Io+mx^2 = mL^2/12 + mL^2/4 = mL^2/3.
T1 = sqrt(I/m•g•x) = sqrt( mL^2/3m•g•x)
=sqrt( L^2/3•g•x) = L/sqrt(3•g•x) =1 s.
L = sqrt(3•g•x),
α =ΔL/(L•ΔT),
ΔL = α •sqrt(3•g•x) • ΔT,
T2 = sqrt ((L+ ΔL)^2/3•g•x) =
=(L+ΔL)/ sqrt(3•g•x)=
= (L + α •sqrt(3•g•x) • ΔT)/ sqrt(3•g•x) =
= ( sqrt(3•g•x) + α •sqrt(3•g•x) • ΔT)/ sqrt(3•g•x)=
= 1+7.8•2.2•10^-5 = 1.000172 s.
b.
Δto= 0.000172 s.
Δt =0.000172•3600•24•14 = 208.0512 s

CAlculating more human .....hope this is making sense to you

Answer 2

The pendulum length gets shorter and the frequency of oscillation increases.

Compute the new length using the coefficent of thermal expansion of steel, 13*10^-6 C^-1.

Call it the new length L2 and the original length L1.

L2/L1 = 1 -13*10^-6*16 = 0.99979

The new period P2 gets multiplied by sqrt(L2/L1)
P2 = 1.0000*0.999896 = 0.999896 s

One week is 10,080 minutes or 6.04800*10^5 s

The new number of oscillations in one week will be 6.04800*10^5/0.999896 = 6.04863*10^5

63 seconds or about one minute will be gained by the clock

HMMMM odd this problem seems to have two answers let me narrow

Answer 3

Your second method is the right way to go.