Question

A clock with a metallic pendulum gains 5 sec each day at a temperature of 15 degree celcious and loses 10 sec each day at a temperature of 30 degree celcious. Find the coeff of thermal expansion of the pendulum metal

Answer 1

in this we have to use mainly three concepts which are  L2=L1(1+(alpha)(change in temprature))

another is binomial theorem which is (1 + x )n= 1 + nx         for x

and the last one which is T = 2\pi\sqrt{\frac{l}{g}}

now it is given that clock is 5 seconds fast each day i.e. 1.16 \times 10-4 seconds

i.e. time period at that temprature is (2 – 1.16 \times 10-4)

and let at \theta temprature it shows correct time i.e. it’s time period is 2 seconds  

now, \frac{T_{1}}{T_{o}}=\sqrt{\frac{L_{1}}{L_{o}}}

and L1 = Lo(1 + \alpha(15  – \theta))

\Rightarrow \frac{T_{1}}{T_{o}}=\sqrt{\frac{L_{o}(1 + \alpha (15 - \theta ))}{L_{o}}}

using binomial theorem

\Rightarrow \frac{T_{1}}{T_{o}}=1 - \frac{1}{2}\alpha(15 - \theta )

\frac{2 - 1.16 \times 10^{-4}}{2} = 1 + \frac{1}{2}\alpha (\theta - 15)

\theta - 15 = \frac{1.16 \times 10^{-4}}{\alpha}        …......................1

similarly  

30 - \theta = \frac{2.32 \times 10^{-4}}{\alpha}   …............................2

now subtracting adding equation 1 and 2 we get

15 = \frac{3.48 \times 10^{-4}}{\alpha}

\Rightarrow \alpha = \frac{3.48 \times 10^{-4}}{15}

substituting it in equation 1 we get  

\theta - 15 = 5

\Rightarrow \theta = 20

so clock will give correct time at 20 oC