Pendulum Clock -- Change in frequency with change in temperature

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  • #1
Sarah0001
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Homework Statement:
For this problem you can consider an ideal pendulum,
for which you will calculate the effect of changing its environment.


(d) Most substances expand with increasing temperature, and so a
metal rod will expand in length by a fraction α δT if the temperature is changed by δT, where α is called the coefficient of linear thermal expansion. Consider the effect of this expansion on a pendulum clock with a pendulum made from brass, with α = 19 × 10^-6 K^-1 .


What temperature change can this clock tolerate if it is to remain accurate to 1 second in 24 hours?
Relevant Equations:
P=2π √(L/g)
1) I do not quite understand how the phrase remain accurate to 1 second in 24 hours? , means ΔP = 1 second,
2) I also don't understand how pendulum period P should be 24 hours
What is the reasoning for both?

The solution is as such

P = 2π √(L/g) P' = 2π √((L+L α δT)/g)

ΔP = P'- P = 2π √((L(1+ α δT)/g) - 2π √(L/g)

where 2π √(L/g) factored out from above to give :

ΔP = 2π √(L/g) ( (√(1+ α δT) - 1)
ΔP = P ( (√(1+ α δT) - 1)

ΔP = 1 sec (which I don't understand this part)
also P = 24 hours = 24*3600 secs ( which I don't understand how one pendulum period should be 24 hours?, I don't understand the reasoning behind this.)

ΔP/P = 1/24*3600
ΔP/P = P ( (√(1+ α δT) - 1) /P

√(1+ α δT) - 1 =1/24*3600

then the value for α can be plugged in and rearranged to find δT.
 
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  • #2
haruspex
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ΔP = 1 sec (which I don't understand this part)
also P = 24 hours = 24*3600 secs ( which I don't understand how one pendulum period should be 24 hours?, I don't understand the reasoning behind this.)
Does it really say ΔP = 1 sec and P = 24 hours, or are you inferring that from:
ΔP/P = 1/24*3600
?
1 second in 24 hours is the fractional change in the clock rate, so is also the fractional change in the period.
 
  • #3
Sarah0001
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Ah yes sorry, I was inferring ΔP = 1 sec and P = 24 hours from the solutions, I just don't understand how they reasoned this in the first place upon reading the question.

(I think this is because do not completely understand part of the question (if the pendulum clock to remain accurate to 1 second in 24 hours?' ) is asking. )
 
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haruspex
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Ah yes sorry, I was inferring ΔP = 1 sec and P = 24 hours from the solutions, I just don't understand how they reasoned this in the first place upon reading the question.
Suppose it should execute N periods in 24 hours. I.e. NP= 24 hours.
If the period has changed to P+ΔP then after N periods the elapsed time is N(P+ΔP). So NΔP= 1 second.
So ΔP/P=(NΔP)/(NP)=1 sec/ 24 hours.
 
  • #5
CPW
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Hi.

I'm thinking the homework problem requires the equation for the period of a physical pendulum (Hint: moment of inertia of a rod about an axis through its end is I= 1/3ML^2). So don't use the equation for the period of the simple pendulum, as you stated in your work.

Perhaps by "ideal" the pendulum swings in a vacuum without the effect of friction, and has small amplitude swings, but is still a rod pendulum.

I hope that helps.
 
  • #6
haruspex
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I'm thinking the homework problem requires the equation for the period of a physical pendulum
No, it changes nothing.
E.g. if the pendulum consists of a point mass M on the end of a uniform rod mass m, length L, the moment of inertia is still proportional to L2, and the torque at a given angle is still proportional to L.
 

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