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Expansion, stress and strain

  • Thread starter ki ki
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  • #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!
 

Answers and Replies

  • #2
Man


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
HallsofIvy
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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
ki ki
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
ki ki
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
Man


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
HallsofIvy
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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
ki ki
alright, i understand now. our prof just gave us that formula last day. thanks for your help!
 

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