91Ó°ÊÓ

1. The pendulum clock is designed to keep accurate time at 23⁰C
2. The rod and the bob of the pendulum clock are made of brass.
3. The coefficient of linear expansion for brass,$\mathrm{Î\pm }=1.9×{10}^{−6}{\text{/}}^{\text{0}}\text{C}$

The period of the pendulum clock depends on the length of the pendulum. Since the rod is made up of metal, it will change in length depending on its temperature. So we determine the change in the length of the rod and then the changed period. A comparison of the new and old periods will give us an idea of the error in the measurement.

Formulae:

Period of oscillation of a pendulum, $T=2\mathrm{Ï€}\sqrt{\frac{L}{g}}$…(¾±)

Where L is the length of the pendulum and $g$is the acceleration due to gravity.

Length of a body due to thermal expansion,$L=L_0(1+\mathrm{Î\pm }\mathrm{Δ}t)$…(¾±¾±)

Where L₀ is the original length of the body, $\mathrm{Î\pm }$is the coefficient of thermal linear expansion of the substance, and $\mathrm{Δ}T$ is the temperature difference at both the ends of the body.

The clock is set to give accurate time at 23⁰C.

But with change in temperature, the length of the rod changes.

Thus, the period of oscillation of the clock can be given using equation (i) in equation (ii) as follows:

$T=2\mathrm{Ï€}\sqrt{\frac{L_0(1−\mathrm{Î\pm }\mathrm{Δ}t)}{g}}$

Let T be the period atstate 0.0⁰C afterbringing it from its state of accurate time23⁰Cand T₀ be the original period before expansion as it operates at this temperature.

Since the temperature is decreasing, we would have value L decreasing. Thus, the above equation becomes

$\begin{array}{c}T=(2\mathrm{Ï€}\sqrt{\frac{L_0}{g}})\sqrt{(1−\mathrm{Î\pm }\mathrm{Δ}t)}\\ =T_0\left(\sqrt{(1−\mathrm{Î\pm }\mathrm{Δ}t)}\right)\text{(fromequation(i))}\\ =T_0{(1−\mathrm{Î\pm }\mathrm{Δ}t)}^{1/2}\end{array}$

We note that $\mathrm{Î\pm }\mathrm{Δ}t≪1$; hence we use the binomial expansion to simplify the above equation as given below (as the pendulum is contracted from a temperature of23⁰Cto0⁰C)

$\begin{array}{c}T=T_0(1−\frac{\mathrm{Î\pm }\mathrm{Δ}t}{2})\text{(neglectingthehighertermsasthevaluegetssmaller)}\\ T=T_0−T_0\frac{\mathrm{Î\pm }\mathrm{Δ}t}{2}\\ \mathrm{Δ}T=−T_0\frac{\mathrm{Î\pm }\mathrm{Δ}t}{2}\text{(}âˆ\mu \mathrm{Δ}T=T−T_0\text{)}\\ =−T_0\left(\frac{19×{10}^{−6}{\text{/}}^0\text{C}×{(0−23)}^0\text{C}}{2}\right)\text{(usingthegivenvalues)}\\ =T_0(2.185×{10}^{−4})\\ =T_0\left(0.99978\right)\\ T−T_0=T_0(2.185×{10}^{−4})\\ T=T_0+T_0(2.185×{10}^{−4})\\ T=T_0(1+2.185×{10}^{−4})\mathrm{....................................}\left(a\right)\end{array}$

Since the multiple factor is greater than 1, the value of T is greater than T₀. It means the clock would run much faster at a temperature of0⁰C.

The magnitude of error can be calculated as

$\begin{array}{c}T=\text{timeof}\text{ }1\text{hr}\\ =3600\text{sec}\end{array}$

From equation (a), the period of oscillations is given by

$$\begin{gathered} T_0=\frac{3600\text{sec}}{\left(1+2.185×{10}^{-4}\right)} \\ =3599.213572\sec \end{gathered}$$

Now, the magnitude of the error in the period of oscillation is given by

$\begin{array}{c}\mathrm{Δ}T=3600\text{s}−3599.213572\text{s}\\ =0.78643\text{s}\\ ≈0.79\text{s/hr}\end{array}$

Hence, the magnitude of error is 0.79 s/hr.