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A second's clock with a iron pendulum is...

A second's clock with a iron pendulum is constructed so a to keep correct time at `10^(@)C`. Give `alpha_("iron")=12xx10^(-6)` per`""^(@)C`. What will be the change in the rate when the temperature rises to `25^(@)C`?

Text Solution

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When the pendulum keeps correct time, its period of vibration is 2 sec and so it makes
`(24xx60xx60)/2=43200` Vibratiions /day
If lenth of the pendulum at `10^(@)C` is `l_(10)` and at `25^(@)C` is `l_(25)`
`:.l_(25)=l_(10)[1+alpha(25-10)]=l_(10)[1+15alpha]`
as `T=2pi sqrt(l//g)` i.e. `T prop sqrt(l)`
i.e. `n prop 1/(sqrt(l))`, n is no of vibrations per sec
`:.(n_(25))/(n_(10))=sqrt((l_(10))/(l_(24)))=[1+15alpha]^(-1//2)~~1=15/2alpha`
`:.n_(25)=n_(10)(1-15/2xx12xx10^(-6))`
`=43200[1-0.00009]=43119.2`
That is the clock makes (3200-43196.12=3.88 vibratiion loss per day. That is clock losses `3.88xx2=7.76` sec per day).
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