How does temperature affect the accuracy of a grandfather clock?

In summary, the conversation discusses how a grandfather's clock is affected by changes in temperature and how to derive an equation to solve for the change in time in a 24 hour period. The period of a pendulum clock is proportional to the square root of the length of the pendulum, and the formula T=2π√(L/G) can be used to calculate the period. The coefficient of expansion, "a", must also be taken into account when calculating the change in length due to temperature. The conversation also touches on the use of differential equations in solving this problem.
  • #1
ki ki
A grandfather's clock is calibrated at a temperature of 20 degrees celcius. The pendulum is a thin brass rod with a heavy mass attached to the end. on a hot day, when the temperature is 30 degrees celcius does the clock 'run' fast or slow? how much time does it gain or lose in a 24 hour period?
I've tried re-arranging delta L = aL delta T so that i could find the length without the change or vice versa and a few other substitution arrangments but nothing seems to be working out for me. please help me with finding how to derive an equation to solve this!
 
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  • #2


All is simple:
T=2 pi radic(L/G)

L=L(0)(1+ alpha*t)
DeltaT=2 pi( radic(L(30)/G)- radic(L(20)/G))
alpha -factor of linear expansion
T-period
L-Length
t-temperature
L(20)-Length if t=20

L(30)-Length if t=30


if DeltaT<0-period is decrease
if DeltaT>0-period is increase
 
  • #3
One of the things you need to know is that the period of a pendulum clock is proportional to the square root of the length of the pendulum. On a "hot day", the metal of the pendulum expands, the pendulum becomes longer, so the period is longer: the clock runs slow.

To get a precise calculation, you will need to know "a", the coefficient of expansion and plug in the change in temperature.
If you take the length of the pendulum at 20 degrees to be L, then when the temperature increases to 30 degrees the pendulum length will increase by aL(30-20)= 10aL so the new length will be L+ 10aL= (1+10a)L. The new period will be [sqrt]((1+10a)L/L)= [sqrt](1+ 10a)
Since the period is "seconds per cycle", it's reciprocal is "cycles per second" = 1/[sqrt](1+10a) and, of course, that is what controls the speed of the clock (one "cycle" of the pendulum might be one second or a part of one second).
 
  • #4
Originally posted by HallsofIvy
One of the things you need to know is that the period of a pendulum clock is proportional to the square root of the length of the pendulum.

how do you know that they are proportional? is there an equation? can you derive it from delta L=aL delta T ?
 
  • #5
All is not quite as simple as it may seem to you, to me:

in radic(L/G) what is 'G'? is that 9.8m/s^2? and where did you get T=2 pi radic(L/G) from? is that for the period of anything or is it specific?
 
  • #6


G-Gravitational constant(9.8m/s^2).
T=2pi(radic(L/G))-the known formula for a mathematical pendulum
Period is time of one fluctuation.
 
  • #7
It's a little harder than that. If you imagine the pendulum at an angle [theta] to the vertical and draw the "force diagram", you find that the net force (gravitational force is downward, tension in the pendulum perpendicular to the motion: break gravitational force into components parallel to and perp to the motion. The two
"perpendicular to motion" forces cancel leaving the force along the line of motion: -mg sin([theta]).

Force= mass*acceleration gives lmd^2s/dt^2= -mg sin([theta]) (s is the arc length along the path of the pendulum bob measured from the bottom.

The general solution to that differential equation is
s(t)= C1 cos([sqrt](g/l)t)+ C2 sin([sqrt](g/l)t) so that the period of the motion is T= 2[pi][sqrt](l/g).

If you are doing a problem like this and haven't taken a course in differential equations, you ought to have been given a formula relating period of the pendulum to it's length.
 
  • #8
alright, i understand now. our prof just gave us that formula last day. thanks for your help!
 

1. What is expansion?

Expansion refers to the increase in size or volume of a material when exposed to heat or pressure. This is due to the increase in thermal energy or molecular movement within the material, causing the atoms to spread out and take up more space.

2. What is stress?

Stress is a force applied to a material that results in deformation or change in shape. It can be caused by external factors such as weight, pressure, or tension, and can also be caused by internal factors such as thermal or chemical changes.

3. What is strain?

Strain is the measure of deformation or change in shape that occurs in a material as a result of stress. It is typically measured as a percentage of the original size or length of the material and can be either elastic (reversible) or plastic (permanent) in nature.

4. How are stress and strain related?

Stress and strain are directly related to each other through a material's elastic modulus, which is a measure of its stiffness or ability to resist deformation. The higher the elastic modulus, the more stress a material can withstand before it reaches its breaking point.

5. How does expansion affect stress and strain?

Expansion can cause stress and strain in a material when it is constrained or unable to move freely. This can lead to thermal stress, which is a type of stress that occurs due to temperature variations within a material. Over time, this can cause fatigue or failure in the material if it is not able to handle the changing temperatures.

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