# Motion of Pendulum Exercise

1. Dec 1, 2007

### BBallman_08

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

An aluminum clock pendulum having a period of 1.00 s keeps perfect time at 20 degrees celcius. (A) When placed in a room at a temperature of -5.0 Celcius, will it gain or lose time? (B)How much time will it gain or lose every hour.

2. Relevant equations

I am pretty sure that this has to due with thermal expansion and motion of a pendulum so:

(A) Delta L = (alpha) (Lo) (delta T)
(B) T = 2pi *sqrt(L/G)

3. The attempt at a solution

I am thinking that the pendulum itself would shrink, therefore making it gain speed, but I just don't know how to set it up correctly??? Any help guys and gals?

Thanks!

2. Dec 1, 2007

### muppet

What is the length of the pendulum once it has changed due to temperature?

3. Dec 1, 2007

### BBallman_08

If I assume length = 1 m, then I come up with a Delta L of .0006 m

4. Dec 1, 2007

### muppet

You don't really need to make assumptions- just do it algebraically.
Then what is the period of your new pendulum?

5. Dec 1, 2007

### BBallman_08

ok, so :

T = (2pi) *Sqrt (.9994/9.8)

t = 2.006 s

So it loses time?

6. Dec 1, 2007

### muppet

Well like I said, I'd do it algebraically (i.e. in terms of L). But that's the next step yes.

7. Dec 1, 2007

### BBallman_08

How would you do it algebraically? Would you solve in terms of Lo ?

8. Dec 1, 2007

### muppet

Just replace 1-0.0006 with Lo-delta L.

9. Dec 1, 2007

### BBallman_08

T = (2pi) *Sqrt (Lo-delta L /9.8) ???

How do I find out how much time it loses? Or would it simply be .0006 sec/min

10. Dec 2, 2007

### muppet

Having read your question again... you know the original period of your pendulum! Divide T' (the period of your new pendulum) by T (the period of your old pendulum) and see what happens. Use no numbers (except 2pi) until you have a formula for T' in terms of T.

11. Dec 3, 2007

### BBallman_08

So would it be :

(2 Pi * sqrt ( l / g)) / 1 sec ?????

12. Dec 3, 2007

### muppet

Think about your equations for T and T'. The ratio T' to T(i.e. T' over 1...) will be equal to the ratio of the RHSs of the two equations.

13. Dec 3, 2007

### BBallman_08

What are you referring to when you say RHSs ?

14. Dec 3, 2007

### muppet

Right Hand Side. Sorry-that's what too much maths does to you...

15. Dec 3, 2007

### BBallman_08

I was getting really confused with the algebra, so I just solved for the original length, Lo. Doing this I found it to be .25 meters.

After that I used the equation :

Delta L = (alpha) (Lo) (delta T), to find the change in length of the aluminum.

Delta L = (24 X 10 ^ -6) ( .25) (-25)

Delta L = -.00015 m

So, the new lenth was .24985 m

I then placed that in the equation for a period and found it to be :

T = 2 pi * sqrt (.24985)/(9.8)

T = 1.003 seconds.

Therefore, the clock slows by .003 seconds per period.

.003 X 3600 s in one hour = Loses 10.8 seconds per hour

Is this correct?

16. Dec 3, 2007

anyone?