Prove that the change in the time period t of a simple pendulum due to a change `DeltaT` of temperature is, `Deltat=1/2alphatDeltaT`, where `alpha` = coefficient of linear expansion.

Explanation

If L is the effective length of a simple pendulum then its time period is, `t=2pisqrt(L/g)`
For a change `DeltaT` of temperature, the length becomes, `L.=L(1+alphaDeltaT)`
Therefore the time period,
`t.=2pisqrt((L.)/g)=2pisqrt((L)/(g)(1+alphaDeltaT))`
`=2pisqrt((L)/(g))(1+alphaDeltaT)^(1//2)=t(1+1/2alphaDeltaT)=t+1/2alphatDeltaT`
[neglecting the terms containing `alpha^(2),alpha^(3)`, etc. since `alpha` is very small]
The change in time period, `Deltat=t.-t=1/2alphatDeltaT` (Proved).
Alternative Method :
`t=2pisqrt(L/g)`
`logt=log2pi+1/2logL-1/2logg`
Differentiating with respect to L,
`1/tdt=1/(2L)dL`
`therefore" "(dt)/t=1/2(dL)/L" "...(1)`
Now `L_(t)=L(1+alphaDeltaT)`
or, `L_(t)-L=LalphaDeltaTor,dL=LalphaDeltaT`
or, `(dL)/L=alphaDeltaT" "...(2)`
Form equation (1) and (2) we get,
`(dt)/t=1/2alphaDeltaTor,dt=1/2alphaDeltaT*t` (Proved).