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?
what is its initial length?
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
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
Your second method is the right way to go.
To find the initial length of the pendulum shaft, we can use the coefficient of linear expansion and the change in temperature.
Let's denote:
- L0 as the initial length of the pendulum shaft
- α as the coefficient of linear expansion of aluminum
- ΔT as the change in temperature from 22 degrees Celsius to 5 degrees Celsius
First, recall that the linear expansion of a material can be given by the formula:
ΔL = α * L0 * ΔT
Here, we know the change in temperature (ΔT = 5°C - 22°C = -17°C) and the coefficient of linear expansion (α = 24.0 * 10^-6 K^-1). We need to find the initial length of the pendulum shaft (L0).
Rearranging the formula, we get:
L0 = ΔL / (α * ΔT)
Now, substituting the values, we have:
L0 = 0 / (24.0 * 10^-6 K^-1 * -17°C)
Note: The change in length (ΔL) is zero since we are finding the initial length of the pendulum shaft.
Simplifying further, we have:
L0 = 0
Therefore, the initial length of the pendulum shaft (L0) is 0.
Since the initial length is 0, we can say that the pendulum shaft is effectively a point mass, and the time period of a pendulum depends only on the length and acceleration due to gravity, and not on the material or the change in temperature. Hence, the question does not provide sufficient information to calculate the time gain or loss in 1 day due to the change in temperature.