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

In summary, the conversation discusses the movement of a grandfather pendulum clock from sea level to a new location 200 m in height. It also mentions the effects of temperature on the period of the clock and provides equations for Newton's gravitational law and the oscillation of a pendulum. The change in period is found to be approximately 2.7 seconds per day, and the necessary change in temperature to maintain precision is determined to be -3.1 K. Alternative methods of finding this answer are also mentioned.
  • #1
Fendergutt
6
0

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.)

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).

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?

Thanks for your help.

 
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  • #2
Or you could just wok out the diffeence in 'g' at the two heights and use T = 2pi sqt(L/g)
 
  • #3


Your approach to the first part of the problem (calculating the change in period due to change in height) seems reasonable. However, it would be good to clarify the values of h and R in the equation you used for the period. It would also be helpful to include the units in your calculations to make it easier to follow.

For the second part of the problem, your approach is correct. The given answer of -3.1 K is likely taking into account the fact that the pendulum is suspended from one end, which would affect the calculation slightly. Another possible explanation could be rounding errors in the given answer. Overall, your method and solution seem sound.
 

1. How does temperature affect the precision of a clock?

As temperature increases, the precision of a clock decreases. This is because temperature changes can cause the material in the clock's mechanism to expand or contract, altering the movement of the clock's hands.

2. Why does the precision of a clock change at different heights above sea level?

The precision of a clock can change at different heights above sea level due to changes in air pressure. As altitude increases, air pressure decreases, which can affect the movement and functioning of a clock's mechanism.

3. Is there an ideal temperature for clock precision?

Yes, most clocks are designed to operate within a specific temperature range for optimal precision. This range is typically between 10-30 degrees Celsius, but can vary depending on the type of clock and its mechanism.

4. How does humidity affect clock precision?

Humidity can also affect clock precision, especially in mechanical clocks. High levels of humidity can cause metal components to rust or become sticky, leading to inaccuracies in timekeeping.

5. Can the precision of a clock be adjusted for different temperatures and heights above sea level?

Yes, some clocks have mechanisms that can be adjusted to compensate for changes in temperature and air pressure at different heights above sea level. However, this may not always be possible for all types of clocks.

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