How to Calculate Error in a Pendulum Clock at Different Temperatures"

[squanto]

I am trying to solve this question..

A pendulum clock with a pendulum made of brass is designed to keep accurate time at 23 °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 know that the the period of oscillation is

T= 2pi * sqrt( L /g)

and that the change in the length of the pendulum
is

-4.37E-4 m

I just don't know how to relate the two to solve
Any help is appreciated.

 

[Tide]

Because T varies as the square root of L the relative errors are related by
[tex]\frac {\delta T}{T} = \frac {1}{2} \frac {\delta L}{L}[/tex]

 

[M98Ranger]

To calculate the error in a pendulum clock at different temperatures, we need to consider the effects of temperature on the length of the pendulum. As stated in the question, the linear expansion coefficient of brass is 19 x 10^-6/C°. This means that for every 1°C increase in temperature, the length of the pendulum will increase by 19 x 10^-6 meters.

To calculate the error at a specific temperature, we can use the formula:

Error (seconds per hour) = (change in length of pendulum / original length of pendulum) * (change in temperature / original temperature) * 3600 seconds

In this case, we are trying to find the error at 0.0°C, so the change in temperature is -23°C (since the original temperature is 23°C). The change in length of the pendulum is -4.37 x 10^-4 meters, as stated in the question.

Substituting these values into the formula, we get:

Error (seconds per hour) = (-4.37 x 10^-4 / original length of pendulum) * (-23 / 23) * 3600 seconds

Now, we need to find the original length of the pendulum. We can use the formula for the period of oscillation:

T = 2π * √(L/g)

Rearranging this formula to solve for L, we get:

L = (T/2π)^2 * g

Since the clock is designed to keep accurate time at 23°C, we can use the original period of oscillation at this temperature, which is 2 seconds. The acceleration due to gravity, g, is a constant of 9.8 m/s^2.

Substituting these values into the formula, we get:

L = (2/2π)^2 * 9.8 m/s^2 = 0.992 m

Now, we can substitute the original length of the pendulum into the formula for error:

Error (seconds per hour) = (-4.37 x 10^-4 / 0.992) * (-23 / 23) * 3600 seconds

Simplifying, we get:

Error (seconds per hour) = -0.784 seconds per hour

Therefore, at 0.0°C, the pendulum clock will slow down by 0.784 seconds per hour