Discussion Forum

A clock pendulum having a coefficient of linear expansion.  $\alpha= 9\times 10^{-7}/^{0}C^{-1}$ has a period of 0.5 s at $20^{0}$ C. If the clock is used in a climate, where the temperature is  $30^{0}$ C, how much time does the clock lose in each oscillation?

Answer:

Explanation:

 Given,  the coefficient of linear expansion,

  $\alpha=9\times10^{-7}{^{o}C^{-1}}$

 Initial time period  , T₀  =0.5 s, initial  temperature T_i= 20° C and final temperature , T_f= 30° C

 Expansion  in length , $\triangle l=l\propto(T_{f}-T_{i})$

$\Rightarrow$    $\triangle l=l\times 9\times 10^{-7}(30-20)$

 Now, the time period of pendulum,

$T=2\pi\sqrt{\frac{l}{g}}$

 error in time period,

$\Rightarrow $    $ \frac{\triangle t}{T}=\frac{1}{2}\frac{\triangle l}{l}+\frac{1}{2}\frac{\triangle g}{g}$

 since, $\triangle g$ =0

$\Rightarrow $    $ \frac{\triangle T}{T}=\frac{1}{2}\frac{\triangle l}{l}$

  Now, substituting  values in the above equation, we get

$\Rightarrow$      $  \frac{\triangle T}{0.5}=\frac{1}{2}\left[\frac{l\times 9 \times 10^{-7}}{l}(30-20)\right]$

$\Rightarrow$    $ \triangle T=\frac{0.5\times 9 \times 10^{-7}\times10}{2}$

$\Rightarrow$    $ \triangle T=2.25 \times 10^{-6}s$