Thermal expansion and keeping time

In summary, the coefficient of thermal expansion for Invar is 0.70*10-6K-1. Given that an Invar pendulum clock keeps perfect time at a room temperature of 20 degrees Celsius, the clock will gain or lose time when the temperature increases to 30 degrees Celsius. This is because the increased temperature causes the pendulum's string length to increase, resulting in a longer period of time for one swing. This means the clock will be slow by T' - T seconds each second, or lose T'-T seconds each second. To calculate the amount of time lost in 24 hours, the equation (T'-T) 24*60*60 can be used, with T' being equal to
  • #1
brad sue
281
0
HI, this problem is supposed to be easy but I don't 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
 
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  • #2
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
Gamma said:
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.


Thank you Gamma,
I understand your explanation.
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 don't know how to get rid of it.

Can you help me on that?
thank you
B
 
  • #4
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
T = 2 pi sqrt(L/g)

T= 1 sec. You can find L
 
  • #6
Gamma said:
T = 2 pi sqrt(L/g)

T= 1 sec. You can find L

Thanks a lot
 
  • #7
Homer - thermal expansion is a fractional amount per degree :
dL/L = coefficient * dT , so the coefficient only has units 1/K .
 

FAQ: Thermal expansion and keeping time

What is thermal expansion?

Thermal expansion is the tendency of matter to increase in volume or size when it is heated and decrease in volume or size when it is cooled.

How does thermal expansion affect objects?

When an object is heated, its molecules vibrate and take up more space, causing the object to expand. When an object is cooled, its molecules move closer together and the object contracts.

What materials are most affected by thermal expansion?

Most materials are affected by thermal expansion to some degree, but materials with higher coefficients of thermal expansion, such as metals and glass, will expand or contract more noticeably with temperature changes.

How is thermal expansion used in everyday life?

Thermal expansion is used in many everyday objects, such as thermometers and thermostats. It is also used in construction, where materials are chosen and designed to account for thermal expansion and contraction.

How does thermal expansion relate to keeping time?

Thermal expansion can affect the accuracy of timekeeping devices, such as clocks and watches. Temperature changes can cause parts of the device to expand or contract, leading to changes in the timekeeping mechanism. This is why many timekeeping devices are designed with materials and mechanisms that are less affected by thermal expansion.

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