# Homework Help: Thermal expansion and keeping time

1. Jan 18, 2006

HI, this problem is supposed to be easy but I dont get it:

The coefficient of thermal expansion for Invar is 0.70*10-6K-1.
( Invar is a steel alloy). Given that an Invar pendulum clock keeps perfect time at a room temperature of 20 degree celcius, how much time will the clockgain or lose per day when it is at room temperature of 30 degree celcius?

can I have some suggestion ?
Thank you

2. Jan 19, 2006

### Gamma

Pendulum period T is a measure of time assuming that the period remains constant. Say one swing of the pendulum is equal to one second. Period of a pendulum is given by T = 2 pi sqrt(L/g)

If the room temp. increased, then the string length is going to increase due to thermal expansion, and as a result, T will increase (say new T is T')

That means, it will take a little longer than one second for your pendulum to make one full swing. This means your pendulum clock is said to be slow. Slow by T' - T seconds each second. Or you loose T'-T seconds each second. So in 24 hours how many seconds would you have lost?

(T'-T) 24*60*60.

T' = 2pi sqrt(L'/g). L' can be found from the given info.

3. Jan 19, 2006

Thank you Gamma,
But I have some issues to find the value of L'= L+dL.

With the data I have I found:
L=dL*(142857.14)

then
L+dL=(1.000007)*L
I still have L as unknow and I dont know how to get rid of it.

Can you help me on that?
thank you
B

4. Jan 19, 2006

### Homer Simpson

Normally thermal expansion is in units length per units temperature.

what is this?: 0.70*10-6K-1.

do you mean .7*10-6 m/K?

5. Jan 19, 2006

### Gamma

T = 2 pi sqrt(L/g)

T= 1 sec. You can find L

6. Jan 20, 2006