Question 961: Thermodynamics & KTG

Question

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

Accepted Answer

Given the problem, we start with the following information:
Coefficient of linear expansion, $ \alpha = 9 	imes 10^{-7} \, ^\circ \mathrm{C}^{-1} $.
Initial period of the pendulum, $ T_0 = 0.5 \, \mathrm{s} $.
Initial temperature, $ T_i = 20^\circ \mathrm{C} $.
Final temperature, $ T_f = 30^\circ \mathrm{C} $.
Calculation Steps:
Determine the expansion in length due to temperature change:
$    \Delta l = l 	imes \alpha 	imes (T_f - T_i)    $
Substituting the values:
$    \Delta l = l 	imes 9 	imes 10^{-7} 	imes (30 - 20)    $
Formulate the time period of the pendulum:
$    T = 2\pi \sqrt{\frac{l}{g}}    $
Find the error in the time period:
$    \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l} + \frac{1}{2} \frac{\Delta g}{g}    $
Note: Since $ \Delta g = 0 $, this simplifies to:
$    \frac{\Delta T}{T} = \frac{1}{2} \frac{\Delta l}{l}    $
Substitute the changes into the equation:
$    \frac{\Delta T}{0.5} = \frac{1}{2} \left( \frac{9 	imes 10^{-7} 	imes 10}{1} ight)    $
Calculate $\Delta T$:
$    \Delta T = \frac{0.5 	imes 9 	imes 10^{-7} 	imes 10}{2}    $
$    \Delta T = 2.25 	imes 10^{-6} \, \mathrm{s}    $
Thus, the clock loses $ 2.25 	imes 10^{-6} $ seconds in each oscillation when the temperature increases to $ 30^\circ \mathrm{C} $.

Answer

$25 	imes 10^{-7} \mathrm{~s}$

Answer

$5 	imes 10^{-7} \mathrm{~s}$

Answer

$1.125 	imes 10^{-6} \mathrm{~s}$

Answer

$2.25 	imes 10^{-6} \mathrm{~s}$

Thermodynamics & KTG

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