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 is reduced by 0.625°C every hour, how much time does the clock gain or lose in 1 day?
what is its initial length?
T = 2 pi sqrt(L/g)= (2 pi/sqrt g)L^.5 = 1 second at 22 deg so you can get initial L
dT/dL = (2 pi/sqrt g).5 /sqrt L
dT/dt = dT/dL * dL/dt
so what is dL/dt
dL/L = 24*10^-6 dT
dL /dT = 24*10^-6 L
dL/dt = dL/dT *dT/dt
= 24*10^-6 (-.625) L
so
dT/dt = [(2 pi/sqrt g).5 /sqrt L][24*10^-6 (-.625) L ]
To find the initial length of the pendulum shaft, we can use the formula for linear expansion:
ΔL = α * L0 * ΔT
Where:
ΔL is the change in length
α is the coefficient of linear expansion (in this case, 24.0 × 10^-6 K^-1)
L0 is the initial length
ΔT is the change in temperature
We are given that the temperature decreases by 0.625°C every hour. In one day, there are 24 hours, so the change in temperature would be:
ΔT = -0.625°C/hour * 24 hours = -15°C
We can now rearrange the formula to solve for the initial length, L0:
L0 = ΔL / (α * ΔT)
Substituting the given values:
L0 = 0 / (24.0 × 10^-6 K^-1 * -15°C)
Note: Since the temperature is decreasing, the change in length ΔL would be zero, as the pendulum shaft contracts and cancels out the expansion.
Therefore, the initial length of the pendulum shaft is zero, assuming no initial expansion occurred.
Now, let's move on to calculating the time gained or lost by the clock in one day.