# Thermal Expansion on Pendum- Period related question

1. Sep 17, 2007

### joeyscl

1. The problem statement, all variables and given/known data

A grandfather clock is calibrated to 20 degrees Celsius. If the room temperature were to be increased to and maintained at 30 degrees, how much would the clock run slower by in a 7 day period? (Yes, the length of the pendulum is NOT given)

Can someone help me with this question please? Also wondering, if you're going to do this question with calculus, is there a way to not use calculus and solve it?

2. Relevant equations

delta T brass= L(initial)x 1.9x10^-5 x 10
period P= 2Pi x Root (L/g)

3. The attempt at a solution
I am completely stumped by this question, all i know is dL/L= 1.9x10^-4

2. Sep 17, 2007

### learningphysics

shouldn't there be an equation relating dL and dt ?

3. Sep 17, 2007

### joeyscl

erm, well, it gives you Period, it also gives u % change in length (well, more like u can calculate it)... how do u go from there?

4. Sep 17, 2007

### genneth

This is a classic problem that needs binomial expansion.

P+dP = 2*pi * sqrt((L+dL)/g)

Expand out the sqrt as a binomial expansion, to first order, then subtract off P to get the value for dP.

5. Sep 17, 2007

### joeyscl

except you're not given L and dL....
you can only calculate dL/L

6. Sep 17, 2007

### joeyscl

cmon, if it were THAT easy i wouldn't be asking for help >,<
I was *easily* one of the best physics students in my Grade in my high school days (only last year, haha) >,<

7. Sep 17, 2007

### genneth

Do you even remember how to do a binomial expansion...? I think you'll find that it just works... Any a word of advice: no matter how good you think you are, you're still here asking questions -- questions that some of the advisers will have been doing without every asking anyone else since long before you were born.

8. Sep 17, 2007

### fikus

genneth si right.
You shold use binomal expansion. The other way of doing it, is to write log of the equation $$P = 2\pi\sqrt{L/g}$$ and then derivate it. What you'll get is
$$\frac{\Delta P}{P}=\frac{1}{2}\frac{\Delta L} {L}$$

9. Sep 20, 2007

### charchar123

hey you're in my phys 153 class eh? lol