How does thermal expansion affect the accuracy of a pendulum clock?

May 26, 2018

the clock will become slower by a small amount due to expansion

Explanation:

A pendulum clock has the Time Period

$T = 2. \pi . \sqrt{\frac{l}{g}}$ where l= length of the pendulum and g the acceleration due to gravity.

By thermal expansion the length of the pendulum will increase say by a small amount $\delta \left(l\right)$

so the new time period  T' = 2.pi. sqrt {( l +delta (l) )/g) }

so, $\frac{T '}{T}$ = $\sqrt{\frac{l + \delta \left(l\right)}{\sqrt{l}}}$

or $\frac{\left(T '\right) - T}{T}$ = $\frac{\sqrt{1 + \delta \left(l\right)} - \sqrt{l}}{\sqrt{l}}$

so the time period of oscillation will increase , thereby the clock will give time as if slow by (T'-T)

In clocks people use seconds pendulum whose time period os 2 secs and length is about 1m and if the length increases to say 1.001 m

then T'- T = 2 s . {sqrt ( 1.001)- 1 }

so the error can be estimated.