Pendulum Clock!

a) From a harmonic oscillator period’s formula solve for the radius.

b) Use the formula for the linear thermal expansion and find the period in each case.

c) In each case find the difference between a second and the period for each pendulum. Consider how many seconds are in a day.

a) The period for a harmonic oscillator is:

\begin{equation*}
T=\frac{1}{2\pi}\sqrt{\frac{L}{g}},
\end{equation*}

where \(L = \ell + r\), and \(\ell = 8r\). Then, by replacing \(L\) in terms of \(r\), and solving for \(r\) we get:

\begin{equation*}
r=\frac{g}{9} \left( \frac{T}{2\pi}\right)^2.
\end{equation*}

With numerical values:

\begin{equation}
r\approx 2.761\,\text{cm}.
\end{equation}

And for \(\ell\):

\begin{equation}
\ell \approx 22.088\,\text{cm}.
\end{equation}

b) The change can be calculated using the linear expansion formula. Explicitly,

\begin{equation*}
L_f=L_0(1+\alpha(T_f-T_0)),
\end{equation*}

which applies for both the length \(\ell\) and the radius \(r\).

For winter, with \(T_f = -10 ^\circ \text{C} \), we get:

\begin{equation*}
\ell_f^{\text{winter}}\approx0.22080\,\text{m},
\end{equation*}

and

\begin{equation*}
r_f^{\text{winter}}\approx0.02760\,\text{m}.
\end{equation*}

The period for winter is:

\begin{equation*}
T^{\text{winter}}\approx 0.99982\,\text{s}.
\end{equation*}

For summer, with \(T_f = 36 ^\circ \text{C} \), we get:

\begin{equation*}
\ell_f^{\text{summer}}\approx0.22093\,\text{m},
\end{equation*}

and

\begin{equation*}
r_f^{\text{summer}}\approx0.02762\,\text{m}.
\end{equation*}

The period for summer is:

\begin{equation*}
T^{\text{summer}}\approx 1.00012\,\text{s}.
\end{equation*}

c) The difference for each period is:

\begin{equation*}
T-T^{\text{winter}}=0.00018\,\text{s},
\end{equation*}

which multiplied by \(86400 \, \text{s}\) we have:

\begin{equation*}
15.552 \, \text{s}.
\end{equation*}

For the other different  we get:

\begin{equation*}
T-T^{\text{summer}}=-0.00012\,\text{s}.
\end{equation*}

which multiplied by \(86400 \, \text{s}\) we have:

\begin{equation*}
-10.368 \, \text{s}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

a) The first part of the problem asks us to find the length of the rod and the diameter of the sphere. Let’s start by making a sketch of the situation described in the problem, as seen in figure 1.

Figure 1: Pendulum clock with a steel rod of length \(\ell\) and a copper sphere of radius \(r\) attached to the rod’s end.

The length of the rod is \(\ell\), and the radius of the sphere is \(r\), so its diameter is \(2r\). From the prompt, we know that the length of the rod is four times the diameter of the sphere; hence, we have the following relation

\begin{equation}
\label{ellr}
\ell=4(2r)=8r.
\end{equation}

From the kinematics of the simple pendulum, we know that the period \(T\), that is, the time it takes to make an oscillation, is given by equation

\begin{equation}
\label{period}
T=\frac{1}{2\pi}\sqrt{\frac{L}{g}},
\end{equation}

where \(L\) is the distance from the pivot point of the pendulum to the center of mass, and \(g\) is the gravitational acceleration of earth.

In our case, because the mass of the rod is negligible, then the center of mass is located in the center of the sphere; thus, the distance from the pivot point to the center of the sphere can be calculated as

\begin{equation}
\label{ELE}
L=\ell+r.
\end{equation}

Using the result of equation \eqref{ELE} into equation \eqref{period}, we obtain

\begin{equation}
\label{period2}
T=2\pi\sqrt{\frac{\ell+r}{g}}.
\end{equation}

The problem indicated that the clock is perfectly synced when the temperature is \(18\,{}^{\circ}\text{C}\), which means that the period is exactly \(1\) second, that is \(T=1\,\text{s}\). Now, we can use the result of equation \eqref{ellr} to replace \(\ell\) in equation \eqref{period2} to get

\begin{equation}
T=2\pi\sqrt{\frac{8r+r}{g}},
\end{equation}

\begin{equation}
\label{period3}
T=2\pi\sqrt{\frac{9r}{g}}.
\end{equation}

Dividing both sides of equation \eqref{period3} by \(2\pi\) and elevating both sides to the 2nd power, we get

\begin{equation}
\left( \frac{T}{2\pi}\right)^2=\frac{9r}{g}.
\end{equation}

Solving for \(r\), we have

\begin{equation}
r=\frac{g}{9} \left( \frac{T}{2\pi}\right)^2.
\end{equation}

Using the numerical values, we obtain

\begin{equation}
r=\frac{9.81\,\text{m/s}^2}{9}\left(\frac{1\,\text{s}}{2\pi}\right)^2,
\end{equation}

\begin{equation}
r\approx 0.02761\,\text{m}=2.761\,\text{cm}.
\end{equation}

Now, using equation \eqref{ellr}, we get the numerical value for \(\ell\), namely

\begin{equation}
\ell=8(0.02761\,\text{m})\approx0.22088\,\text{m}=22.088\,\text{cm}.
\end{equation}

b) We need now to find the period duration for the winter and summer months. Due to the change in temperature during these seasons, the length of the rod changes, and the radius of the sphere also changes. The change can be calculated using the linear expansion formula. Explicitly,

\begin{equation}
\label{explineal}
L_f=L_0(1+\alpha(T_f-T_0)),
\end{equation}

where \(L_f\) is the length of the object at temperature \(T_f\), \(L_0\) is the length of the object at temperature \(T_0\) and \(\alpha\) is the linear thermal expansion coefficient, that only depends on the material. We can then use the relation given in equation \eqref{explineal} to write the expressions for the expansion of \(\ell\) and \(r\). Explicitly,

\begin{equation}
\label{expell}
\ell_f=\ell_0(1+\alpha_{\text{steel}}(T_f-T_0)),
\end{equation}

and

\begin{equation}
\label{expr}
r_f=r_0(1+\alpha_{\text{copper}}(T_f-T_0)),
\end{equation}

where we have used different \(\alpha\) depending on the material of each object. From part (a), we know the length \(\ell\) and the radius \(r\) for a temperature of \(18\,{}^{\circ}\text{C}\). Therefore, \(T_0=18\,{}^{\circ}\text{C}\), \(\ell_0=0.22088\,\text{m}\) and \(r_0=0.02761\,\text{m}\).

Then for winter \(T_{f}=-10\,{}^{\circ}\text{C}\), we can use expressions \eqref{expell} and \eqref{expr} to calculate numerically the length at such temperature:

\begin{equation}
\ell_f^{\text{winter}}=(0.22088\,\text{m})(1+(12\times10^{-6}\,{}^{\circ}\text{C}^{-1})(-10\,{}^{\circ}\text{C}-18\,{}^{\circ}\text{C})),
\end{equation}

\begin{equation}
\label{ellwinter}
\ell_f^{\text{winter}}\approx0.22080\,\text{m},
\end{equation}

and

\begin{equation}
r_f^{\text{winter}}=(0.02761\,\text{m})(1+(17\times10^{-6}\,{}^{\circ}\text{C}^{-1})(-10\,{}^{\circ}\text{C}-18\,{}^{\circ}\text{C})),
\end{equation}

\begin{equation}
\label{rwinter}
r_f^{\text{winter}}\approx0.02760\,\text{m}.
\end{equation}

Then using equation \eqref{period2} and the results from equations \eqref{ellwinter} and \eqref{rwinter}, we can calculate the period for winter as

\begin{equation}
T^{\text{winter}}=2\pi\sqrt{\frac{\ell_f^{\text{winter}}+r_f^{\text{winter}}}{g}},
\end{equation}

then, numerically, we get

\begin{equation}
T^{\text{winter}}=2\pi\sqrt{\frac{0.22080\,\text{m}+0.02760\,\text{m}}{9.81\,\text{m/s}^2}},
\end{equation}

\begin{equation}
T^{\text{winter}}\approx 0.99982\,\text{s}.
\end{equation}

Thus, the period in winter is less than in spring or autumn.

Now, let’s repeat the same analysis for summer, where the temperature is \(T_f=38\,{}^{\circ}\text{C}\). We can then use expressions \eqref{expell} and \eqref{expr} to calculate numerically the length at such temperature:

\begin{equation}
\ell_f^{\text{summer}}=(0.22088\,\text{m})(1+(12\times10^{-6}\,{}^{\circ}\text{C}^{-1})(38\,{}^{\circ}\text{C}-18\,{}^{\circ}\text{C})),
\end{equation}

\begin{equation}
\label{ellsummer}
\ell_f^{\text{summer}}\approx0.22093\,\text{m},
\end{equation}

and

\begin{equation}
r_f^{\text{summer}}=(0.02761\,\text{m})(1+(17\times10^{-6}\,{}^{\circ}\text{C}^{-1})(38\,{}^{\circ}\text{C}-18\,{}^{\circ}\text{C})),
\end{equation}

\begin{equation}
\label{rsummer}
r_f^{\text{summer}}\approx0.02762\,\text{m}.
\end{equation}

Then, using equation \eqref{period2} and the results from equations \eqref{ellwinter} and \eqref{rwinter}, we can calculate the period for winter as

\begin{equation}
T^{\text{summer}}=2\pi\sqrt{\frac{\ell_f^{\text{summer}}+r_f^{\text{summer}}}{g}},
\end{equation}

then, numerically, we get

\begin{equation}
T^{\text{summer}}=2\pi\sqrt{\frac{0.22093\,\text{m}+0.02762\,\text{m}}{9.81\,\text{m/s}^2}},
\end{equation}

\begin{equation}
T^{\text{summer}}\approx 1.00012\,\text{s}.
\end{equation}

Thus, the period in summer is greater than in spring or autumn.

c) We need to find out how much time the clock drifts during the winter and summer months. To do this, we need to find the difference between the period in sync and the period during winter or summer. Then, we multiply this difference by the number of seconds in one day. Let’s begin by calculating the number of seconds in one day: we multiply the number of seconds in one minute by the number of minutes in an hour by the number of hours in a day, obtaining

\begin{equation}
60\times60\times24=86400\, \text{seconds in a day}
\end{equation}

Now, for winter the difference in time is

\begin{equation}
T-T^{\text{winter}}=1\,\text{s}-0.99982\,\text{s}=0.00018\,\text{s}.
\end{equation}

Which, when multiplied by the number of seconds in a day, gives us

\begin{equation}
0.00018\,\text{s}\times 86400=15.552\,\text{s}.
\end{equation}

Thus, in winter, the clock drifts forward an amount of approximately 15 seconds per day. In the case of summer, the difference in time is

\begin{equation}
T-T^{\text{summer}}=1\,\text{s}-1.00012\,\text{s}=-0.00012\,\text{s}.
\end{equation}

Which, when multiplied by the number of seconds in a day, gives us

\begin{equation}
-0.00012\,\text{s}\times 86400=-10.368\,\text{s}.
\end{equation}

Thus, in summer, the clock drifts backward an amount of approximately 10 seconds per day.

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