Thermal expansion and a pendulum clock

[jealey]

I need a little help. The problem is as follows:


A pendulum clock with a pendulum made of brass is designed to keep accurate time at 18 °C. If the clock operates at 0.0°C, what is the magnitude of its error, in seconds per hour (use a minus sign to indicate slowing down)? The linear expansion coefficient of brass is 19 x 10-6 /C°.


I tried multiplying the change in temperature by the coefficient of brass, but that does not come out right. I also tried multiplying that answer by 3600 to get the seconds but that was not right either. What else is there to do?

 

[Tide]

I need a little help. The problem is as follows:


A pendulum clock with a pendulum made of brass is designed to keep accurate time at 18 °C. If the clock operates at 0.0°C, what is the magnitude of its error, in seconds per hour (use a minus sign to indicate slowing down)? The linear expansion coefficient of brass is 19 x 10-6 /C°.


I tried multiplying the change in temperature by the coefficient of brass, but that does not come out right. I also tried multiplying that answer by 3600 to get the seconds but that was not right either. What else is there to do?

Did you use the fact that the period of the pendulum is proportional to $\sqrt L$?

 

[M98Ranger]

Thermal expansion can definitely have an effect on the accuracy of a pendulum clock. As the temperature changes, the length of the pendulum will also change due to the expansion of the brass material. This change in length can cause the clock to either run faster or slower, depending on the direction of the change in temperature.

To calculate the magnitude of the error in seconds per hour, we need to use the formula for thermal expansion: ΔL = αLΔT, where ΔL is the change in length, α is the linear expansion coefficient, L is the original length, and ΔT is the change in temperature.

In this case, we know that the linear expansion coefficient for brass is 19 x 10^-6 /C°, and that the clock is designed to keep accurate time at 18 °C. So, if the clock operates at 0.0°C, we can calculate the change in temperature as -18°C (since the temperature is decreasing). We also know that the length of the pendulum is constant, so we can use the original length for L.

Now, we can plug in these values into the formula: ΔL = (19 x 10^-6 /C°) x (L) x (-18°C). This gives us the change in length of the pendulum due to the change in temperature. We can then convert this into seconds per hour by multiplying it by 3600 (since there are 3600 seconds in an hour).

So, the magnitude of the error in seconds per hour would be (19 x 10^-6 /C°) x (L) x (-18°C) x 3600. This will give us the number of seconds that the clock will either gain or lose in an hour at 0.0°C. Remember to include the minus sign to indicate that the clock will slow down at this temperature.

I hope this helps! Keep in mind that there may be other factors that can affect the accuracy of a pendulum clock, such as air resistance and friction, so this calculation may not be completely accurate. But, it can give us a general idea of the effect of thermal expansion on the clock.