# Clock precision as a function of temperature and height above sea level

#### Fendergutt

1. Homework Statement

A grandfather pendulum clock is to be moved from a location at sea level to a new location approximately 200 m up in height.

We assume that the Earth's radius is 6400 km. We know Newton's gravitational law and the oscillation equation for a pendulum.

(*) How much will the period of the clock increase, expressed as seconds/day as it is moved from the old height to the new height?

(**) We also know that the pendulum's period will change with temperature. If the thermal expansion coefficient for the pendulum is α = 2.03E-5 per Kelvin, how much do we have to alter the room temperature from the usual temperature in order for the clock to be precise again? (You may assume that the pendulum has the form of a rod, and that it is suspended from one end.) [Answer ΔT ~ -(1/α)*((2h)/R) ~ -3.1 K] (The difference between these given variables h and R are not clear from the text.)

2. Homework Equations

Newton's grav. law: F = G*(m1*m2)/r²
Oscillation equation (I combined relations found in the book by Tipler and Mosca): T = (2π)/ω. ω = sqrt(k/m).

3. The Attempt at a Solution

(*) I imagined two extremely long pendulum clocks with extremely long periods:

F = G*(m_1m_2)/r² = m_1*a [m_1 is mass of pendulum]

a = G*m_2/r² [m_2 is mass of Earth]

m = m_2

T = 2π * sqrt(m_2/a) = 2πsqrt(r²/G) [G gravitational constant]

--> T_1 = 2π*(6400E3)*sqrt(1/6.67E-11) ~ 4.923760562E12, ie. seconds

T_2 = 2π*(6400E3+200)*sqrt(1/6.67E-11) ~ 4.923914429E12

Difference in period, in seconds/day:

(T_2-T_1)/T_2 * 60²*24 s/d ~ 2.7 s/d

which was also given as correct in the paper. Please give comments on this anyway if you think my assumptions / method of solution is strange etc.

(**)

I found the equation ΔL/L = αΔT <--> ΔT = ΔL/(α*L) = 200 / (6400E3 * 2.03E-5 K-1) ~ 1.539 K

The given answer is as mentioned above -3.1 K. I observe that 1.539 K * 2 ~ 3.1 K. Should I use some other relations?

1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution
1. Homework Statement

2. Homework Equations

3. The Attempt at a Solution

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#### mgb_phys

Homework Helper
Or you could just wok out the diffeence in 'g' at the two heights and use T = 2pi sqt(L/g)

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