# A pendulum clock keeps correct time at 0^oC. If the coefficient of linear expansion is \alpha,...

## Question:

A pendulum clock keeps correct time at {eq}0^oC {/eq}. If the coefficient of linear expansion is {eq}\alpha {/eq}, then what will be the loss in time per day, when the temperature rises by {eq}t^oC {/eq}?

## Time Loss In a Pendulum Clock

In a pendulum clock, the time period of the oscillation of the pendulum is one second. The time period is a function of the length of the pendulum. The longer the pendulum is the more will be the time period. The length is, on the other hand, a function of the ambient temperature. The higher the temperature the more will be the length. Thus the time period has an indirect dependence on the surrounding temperature.

If a rod of length {eq}\displaystyle {l_0} {/eq} at a temperature of {eq}\displaystyle {0^{\circ}\ \text{C}} {/eq} is heated to a temperature of {eq}\displaystyle {t^{\circ}\ \text{C}} {/eq} then the length {eq}\displaystyle {l} {/eq} is given by,

{eq}\displaystyle {l=l_0(1+\alpha t)} {/eq}.

Here {eq}\displaystyle {\alpha} {/eq} is the temperature coefficient of linear expansion.

The time period of oscillation of a pendulum of length {eq}\displaystyle {l} {/eq} is,

{eq}\displaystyle {T=2\pi\times \sqrt{\dfrac{l}{g}}=2 \pi \sqrt{\dfrac{l_0(1+\alpha t)}{g}}} {/eq}

{eq}\displaystyle {T=2\pi \sqrt{\dfrac{l_0}{g}}\times \left(1+\alpha t \right)^{\dfrac{1}{2}}} {/eq}

Now we can use the binomial approximation,

{eq}\displaystyle {(1+x)^n\approx (1+nx)} {/eq} if {eq}\displaystyle {|x|} {/eq} is very small compared to 1.

Then,

{eq}\displaystyle {T\approx T_0 (1+\dfrac{1}{2}\alpha t)} {/eq}

Here {eq}\displaystyle {T_0=2\pi \sqrt{\dfrac{l}{g}}} {/eq}

Hence,

{eq}\displaystyle {\Delta T=T-T_0 \approx T_0\times \dfrac{1}{2} \alpha t} {/eq}

Or,

{eq}\displaystyle {\dfrac{\Delta T}{T_0}=\dfrac{1}{2}\alpha t} {/eq}

This gives the time lost per oscillation. Now one oscillation corresponds to one second. In a day there are 86400 seconds. Hence the time lost in a day will be,

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

Or,

{eq}\displaystyle {\boxed{\Delta T=43200 \alpha t}} {/eq}