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Effect of Temperature on the Time Period of a Simple Pendulum

In this article we are going to discuss the effect of temperature on the time period of a simple pendulum, i.e., how the time read by a pendulum clock is affected by the rise and fall in temperature.

A pendulum clock consists of a metal bar attached with a bob at one end and fixed at the other end.

The length of pendulum (length of metal bar) depends on temperature, and hence time period of clock depends on temperature.

Let a pendulum clock read correct time when its length is l₀. Its time period t is given by :-

$\displaystyle t=2\pi \sqrt{\frac{{{l}_{0}}}{g}}$

Now suppose that the temperature is raised by ΔT, then new time period becomes

$\displaystyle {{t}^{'}}=2\pi \sqrt{\frac{l}{g}}$

$\displaystyle \Rightarrow \frac{{{t}^{'}}}{t}=\sqrt{\frac{l}{{{l}_{0}}}}$

Now from thermal expansion, new length of pendulum(l), is given by :-

l = l₀( 1 + α ΔT)

Hence,

$\displaystyle \Rightarrow \frac{{{t}^{'}}}{t}=\sqrt{\frac{{{l}_{0}}(1+\alpha \Delta T)}{{{l}_{0}}}}$

$\displaystyle \Rightarrow \frac{{{t}^{'}}}{t}=\sqrt{(1+\alpha \Delta T)}={{(1+\alpha \Delta T)}^{1/2}}$

$\displaystyle \Rightarrow \frac{{{t}^{'}}}{t}=1+\frac{1}{2}\alpha \Delta T$

[As the value of α is very small]

$\displaystyle \Rightarrow \frac{{{t}^{'}}}{t}-1=\frac{1}{2}\alpha \Delta T$

$\displaystyle \Rightarrow \frac{{{t}^{'}}-t}{t}=\frac{1}{2}\alpha \Delta T$

Now t’-t = Δt, is loss in time

So,

Fractional loss in time $\displaystyle \Rightarrow \frac{\Delta t}{t}=\frac{1}{2}\alpha \Delta T$

Here fractional loss in time means, per second loss in time.

Number of seconds in one day = 86400

⇒ Time lost per day(Δt) = $\displaystyle \frac{1}{2}\alpha \Delta T\times 86400\sec.$

Example 1. A pendulum clock gives correct time at $20º C$ at a place where $g = 9.800$ $m$ $s^{-2}$ . The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where $g = 9.788$ $m$ $s^{-2}$ . At what temperature will it give correct time? Coefficient of linear expansion of steel $= 12 \times 10^{-6º} C^{-1}$ .

Solution.

As, $\displaystyle t=2\pi \sqrt{\frac{l}{g}}$

For the pendulum to keep correct time, its time period must be 2seconds.

⇒ $\displaystyle 2=2\pi \sqrt{\frac{{{l}_{0}}}{9.8}}$ …..(1)

and at another place

⇒ $\displaystyle 2=2\pi \sqrt{\frac{{{l}_{0}}[1+12\times {{10}^{-6}}\times (T-20)]}{9.788}}$ …..(2)

Equating (1) and (2), we get

$\displaystyle \frac{{{l}_{0}}}{9.8}=\frac{{{l}_{0}}}{9.788}[1+12\times {{10}^{-6}}\times (T-20)]$

$\displaystyle \Rightarrow \frac{9.788}{9.8}=1+12\times {{10}^{-6}}\times (T-20)$

$\displaystyle \Rightarrow \frac{9.788}{9.8}-1=12\times {{10}^{-6}}\times (T-20)$

$\displaystyle \Rightarrow -\frac{0.012}{9.8}=12\times {{10}^{-6}}\times (T-20)$

$\displaystyle \Rightarrow -\frac{12\times {{10}^{-3}}}{9.8}=12\times {{10}^{-6}}\times (T-20)$

$\displaystyle \Rightarrow -\frac{1000}{9.8}=T-20$

$\displaystyle T-20=-102$

⇒ T = -82ºC

Example 2. A pendulum clock consists of an iron rod connected to a small, heavy bob. If it is designed to keep correct time at 20°C, how fast or slow will it go in 24 hours at 40°C ? Coefficient of linear expansion of iron = 1.2 × 10^–6/°C.

Solution.

Time lost per day(Δt) = $\displaystyle \frac{1}{2}\alpha \Delta T\times 86400\sec.$

$\displaystyle \Delta t=\frac{1}{2}\times 1.2\times {{10}^{-6}}\times 20\times 86400$

Δt = 1.04 sec

This is loss of time as final temperature(40°C) is greater than initial temperature(20°C). As the temperature increases, the time period also increases. Thus, the clock goes slow.

Next Topic :- Error in Metal Scale Reading Due To Expansion and Contraction

Previous Topic :- Relation Between alpha beta and gamma

Effect of Temperature on the Time Period of a Simple Pendulum

Effect of Temperature on the Time Period of a Simple Pendulum

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