 #1
Sarah0001
 31
 1
 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.
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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