A clock with a metallic pendulum gains 5 \ s each day at a temperature of 15^oC and loses 10 \ s...

Question:

A clock with a metallic pendulum gains {eq}5 \ s {/eq} each day at a temperature of {eq}15^oC {/eq} and loses {eq}10 \ s {/eq} each day at a temperature of {eq}30^oC {/eq}. Find the coefficient of thermal expansion of the pendulum metal.

Thermal Expansion:

The value of thermal expansion of a metal can be evaluated by examining the original length of the object, the change in the temperature of the metal and the thermal coefficient of expansion of that particular metal.

Answer and Explanation:


Given data:

  • Gain in time, {eq}d{T_1} = 5\;{\rm{s/day}} {/eq}
  • Temperature, {eq}{t_1} = 15{\rm{^\circ C}} {/eq}
  • Loss in time, {eq}d{T_2} = 10\;{\rm{s/day}} {/eq}
  • Temperature, {eq}{t_2} = 30{\rm{^\circ C}} {/eq}


The expression for the time period of a pendulum is,

{eq}T = 2\pi \sqrt {\dfrac{l}{g}} {/eq}

This can be written as,

{eq}{T^2} = 4{\pi ^2}\left( {\dfrac{l}{g}} \right).......................(1) {/eq}

Differentiate the above equation,

{eq}2TdT = \dfrac{{4{\pi ^2}}}{g}\left( {dl} \right).....................(2) {/eq}

Also the equation (1) can be written as,

{eq}\dfrac{{{T^2}}}{l} = \dfrac{{4{\pi ^2}}}{g} {/eq}

Substitute this value in equation (2)

{eq}\begin{align*} 2TdT &= \left( {\dfrac{{{T^2}}}{l}} \right)dl\\ \dfrac{{dT}}{T} &= \dfrac{1}{2}\dfrac{{dl}}{l} = \dfrac{{1l\alpha \Delta t}}{{2l}} = \dfrac{1}{2}\alpha \Delta t\\ dT &= \dfrac{1}{2}\alpha \Delta t \times T \end{align*} {/eq}

Here, {eq}T = 1\;{\rm{day}} = 86400\;{\rm{s}} {/eq}

{eq}dT = \dfrac{1}{2}\alpha \Delta t \times 86400 {/eq}

At {eq}15{\rm{^\circ C}} {/eq} the gain is,

{eq}\begin{align*} d{T_1} &= \dfrac{1}{2}\alpha \left( {t - {t_1}} \right) \times 86400\\ 5 &= \dfrac{\alpha }{2}\left( {t - 15} \right) \times 86400.......................(3) \end{align*} {/eq}

Here, {eq}t {/eq} is the temperature at which the clock gives the correct time.

At {eq}30{\rm{^\circ C}} {/eq} the loss is,

{eq}\begin{align*} d{T_2} &= \dfrac{1}{2}\alpha \left( {{t_2} - t} \right) \times 86400\\ 10 &= \dfrac{\alpha }{2}\left( {30 - t} \right) \times 86400......................(4) \end{align*} {/eq}

Now divide the equation (4) by equation (3)

{eq}\begin{align*} \dfrac{{10}}{5} &= \dfrac{{\dfrac{\alpha }{2}\left( {30 - t} \right) \times 86400}}{{\dfrac{\alpha }{2}\left( {t - 15} \right) \times 86400}}\\ 2 &= \dfrac{{30 - t}}{{t - 15}}\\ t &= 20{\rm{^\circ C}} \end{align*} {/eq}

Substitute this value in equation (3)

{eq}\begin{align*} 5 &= \dfrac{\alpha }{2}\left( {t - 15} \right) \times 86400\\ 5 &= \dfrac{\alpha }{2}\left( {20 - 15} \right) \times 86400\\ 10 &= \alpha \left( {20 - 15} \right) \times 86400\\ \alpha &= 2.31 \times {10^{ - 5}}\;{\rm{^\circ }}{{\rm{C}}^{{\rm{ - 1}}}} \end{align*} {/eq}

Therefore, the coefficient of thermal expansion of the pendulum metal is {eq}2.31 \times {10^{ - 5}}\;{\rm{^\circ }}{{\rm{C}}^{{\rm{ - 1}}}} {/eq} .


Learn more about this topic:

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Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12
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