Clock Pendulum Bob
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Pendulum clocks, thermal expansion and time?
A particular pendulum clock keeps correct time at 20.0 degrees celsuis The pendulum shaft is steel, and its mass can be ignored compared with that of the bob.
How many seconds per day will the clock gain or lose at 10.0 degrees Celsius.
given the coefficient of linear expansion of steel is 1.2*10^-5
would the period of the pendulum also affect time?
thanks, 1st year university
With regards to your question..
"would the period of the pendulum also affect time?"
the answer is absolutely. If fact it is the period changing that effects the time.
First T=2pi*sqrt(L/g) (T is the period , g=9.81 m/s/s)
Since the length of the pendulum changes
let this be the equation before..
T1=2pi*sqrt(L1/g)
and this after
T2=2pi*sqrt(L2/g)
To simplify the below let me replace the number 1.2*10^-5 with the letter k.
Note that k is the ratio of the change of length for each degree of temperature change. ie Delta L = L * k * Delta Temperature.
Because the pendulum is colder the shaft length will shrink.
If it was L1 before it will be L2= L1-L1 * k * 10 after.
The above simplifies to L2=L1(1-k*10)
Now we find the ratio of the period before and after.
T2/T1= 2pi*sqrt(L2/g) / 2pi*sqrt(L1/g)
which equals
T2/T1= 2pi*sqrt(L1(1-k*10)/g) / 2pi*sqrt(L1/g)
which equals
T2/T1= sqrt(L1(1-k*10)/g) / sqrt(L1/g)
which equals
T2/T1= sqrt( (1-k*10) )
The ratio of the two periods equals the ratio of times that the hour hand takes to go around once for example.
Therefore
Time2/Time1=T2/T1= sqrt( (1-k*10) )
let time1 be the number of seconds in day and you have your answer.
delta time =Time1-Time2
That was a pretty hard question.
Could you tell me what grade level you were asked this question in?
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