Oscillation Frequency

The equivalent system for a spring is a capacitor, thus the equivalent value for series and parallel connection can be found exactly the same. Then, use the equation for the frequency with the found spring constant.

For a series connection we have:

\begin{equation*}
\frac{1}{k_{\text{series}}}=\frac{1}{k}+\frac{1}{k'},
\end{equation*}

and for parallel's connections we have:

\begin{equation*}
k_{\text{parallel}}=k+k'.
\end{equation*}

The two springs of constant \(k_2\) are in series. Then:

\begin{equation*}
k_{\text{series}}=\frac{k_2}{2}.
\end{equation*}

Now we can reduce the springs of constant \(k_1\) which are in parallel:

\begin{equation*}
k_{\text{parallel}}=2k_1.
\end{equation*}

The springs of constant \(k_3\) are in parallel. So:

\begin{equation*}
k_{\text{parallel}}=2k_3.
\end{equation*}

The following reduction is:

\begin{equation*}
k_{\text{parallel}}=2k_1+\frac{k_2}{2}.
\end{equation*}

Finally, the resulting springs are in series. Then:

\begin{equation*}
k_{\text{series}}=\left(\frac{1}{2k_1+\frac{k_2}{2}}+\frac{1}{2k_3}\right)^{-1}.
\end{equation*}

Since the angular frequency \(\omega = \sqrt{\frac{k}{m}}\) and the frequency \(f = \frac{\omega}{2 \pi}\). Then:

\begin{equation*}
f=\frac{1}{2\pi}\sqrt{\frac{\left(\frac{1}{2k_1+\frac{k_2}{2}}+\frac{1}{2k_3}\right)^{-1}}{m}}.
\end{equation*}

Using the numerical values we get:

\begin{equation*}
f\approx 0.58\,\text{Hz}.
\end{equation*}

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

We need to find the system frequency of oscillation, when the mass is free to oscillate. In order to solve this problem, we must reduce it to a problem with a mass and one spring, where the value of its constant \(k\) will be the equivalent of the array shown in the first figure. To make this reduction we must identify which springs are in series and which ones are in parallel and how to find an equivalent in both cases.

We say that two springs are in series when they are connected by one end exclusively, as seen in figure 1.

\begin{figure}[H] \centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/springseries.png}
\caption{Two springs in series}
\label{fig:springseries}
\end{figure}

Suppose the springs in figure have constants \(k\) for the spring in the right and \(k'\) for the spring in the left. When a force \(F\) is applied to the springs on the right, it is transmitted in the same magnitude to the spring to the left. The deformation of the spring on the right being \(x\) and on the left \(x'\). The total deformation of the array of springs is

\begin{equation}
\label{xtotal}
x_{\text{total}}=x+x'.
\end{equation}

Using Hooke's law concentrating in just the magnitudes

\begin{equation}
\label{hooke1}
F=kx
\end{equation}

to solve for \(x\) we get

\begin{equation}
\label{hooke2}
x=\frac{F}{k}.
\end{equation}

Using the expression given in equation \eqref{hooke2} to express \(x\) and \(x'\) in equation \eqref{xtotal} we obtain

\begin{equation}
x_{\text{total}}=\frac{F}{k}+\frac{F'}{k'}.
\end{equation}

Because the forces on each spring are equal then \(F'=F\), thus we can write

\begin{equation}
x_{\text{total}}=\frac{F}{k}+\frac{F}{k'},
\end{equation}

which after factorizing \(F\) is

\begin{equation}
\label{xtotal2}
x_{\text{total}}=F\left(\frac{1}{k}+\frac{1}{k'}\right).
\end{equation}

If we define an equivalent constant \(k_{\text{series}}\) as

\begin{equation}
\label{kseries}
\frac{1}{k_{\text{series}}}=\frac{1}{k}+\frac{1}{k'},
\end{equation}

then equation \eqref{xtotal2} can be written as

\begin{equation}
x_{\text{total}}=\frac{F}{k_{\text{series}}},
\end{equation}

which is equivalent to Hooke's law in its form of equation \eqref{hooke2} with an equivalent constant \(k_{\text{series}}\) defined by equation \eqref{kseries}. Therefore, the system of two springs with constants \(k\) and \(k'\) can be replaced by one spring of constant \(k_{\text{series}}\) (from equation \eqref{kseries})

\begin{equation}
\label{kseries1}
\frac{1}{k_{\text{series}}}=\left(\frac{1}{k}+\frac{1}{k'}\right).
\end{equation}

We can make a similar analysis for springs in parallel. We say that two springs are in parallel when the are connected by both ends, as seen in Figure the previous figure.

\begin{figure}
\centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/springsparalell.png}
\caption{Two springs in parallel}
\label{fig:springparallel}
\end{figure}

Suppose the springs in the third figure have constants \(k\) for the spring in the top and \(k'\) for the spring in the bottom. When a force \(F\) is applied the right, part of this force is transmitted to the spring on the top and part to spring in the bottom. The deformation of both springs will be the same being the one from the top \(x\) and on the bottom \(x'\). The total force \(F\) applied to the array of springs is

\begin{equation}
\label{Ftotal}
F_{\text{total}}=F+F,
\end{equation}

where \(F\) is the force exerted on the top spring and \(F'\) is the force exerted on the bottom spring.

Using the expression given in equation \eqref{hooke1} to express \(F\) and \(F'\) in equation \eqref{Ftotal} we obtain

\begin{equation}
F_{\text{total}}=kx+k'x'.
\end{equation}

Because the deformation of both springs are equal then \(x'=x\), thus we can write

\begin{equation}
F_{\text{total}}=kx+k'x,
\end{equation}

which after factorizing \(x\) is

\begin{equation}
\label{Ftotal2}
F_{\text{total}}=(k+k')x
\end{equation}

If we define an equivalent constant \(k_{\text{parallel}}\) as

\begin{equation}
\label{kparallel}
k_{\text{parallel}}=k+k',
\end{equation}

then equation \eqref{Ftotal2} can be written as

\begin{equation}
F_{\text{total}}=k_{\text{parallel}}x
\end{equation}

which is equivalent to Hooke's law in its form of equation \eqref{hooke1} with an equivalent constant \(k_{\text{parallel}}\) defined by equation \eqref{kparallel}. Therefore, the system of two springs with constants \(k\) and \(k'\) can be replaced by one spring of constant \(k_{\text{parallel}}\) given by equation \eqref{kparallel}.

We can now use equations \eqref{kseries1} and \eqref{kparallel} to reduce our array of springs. The two springs of constant \(k_2\) are clearly in series, then they can be replaced by one spring of equivalent constant (according to equation \eqref{kseries1})

\begin{equation}
k_{\text{series}}=\left(\frac{1}{k_2}+\frac{1}{k_2}\right)^{-1},
\end{equation}

\begin{equation}
k_{\text{series}}=\left(\frac{2}{k_2}\right)^{-1},
\end{equation}

\begin{equation}
k_{\text{series}}=\frac{k_2}{2}.
\end{equation}

We can then draw the array again as in the following figure.

\begin{figure}[H] \centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/array2.png}
\caption{Array after reduction}
\label{fig:array2}
\end{figure}

Now we can reduce the springs of constant \(k_1\) which are in parallel using equation \eqref{kparallel} to get

\begin{equation}
k_{\text{parallel}}=k_1+k_1,
\end{equation}

\begin{equation}
k_{\text{parallel}}=2k_1.
\end{equation}

Then we can redraw the array of springs as seen in the next figure.

\begin{figure}[H] \centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/array3.png}
\caption{Array after reduction}
\label{fig:array3}
\end{figure}

Let us now reduce the springs of constant \(k_3\) which are in parallel using equation \eqref{kparallel} to get

\begin{equation}
k_{\text{parallel}}=k_3+k_3,
\end{equation}

\begin{equation}
k_{\text{parallel}}=2k_3.
\end{equation}

Then we can redraw the array of springs as seen in the next figure.

\begin{figure}[H] \centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/array4.png}
\caption{Array after reduction}
\label{fig:array4}
\end{figure}

We can further reduce the array of spring if we consider that the two left-most spring are in parallel. Then they can be reduced to one spring using equation \eqref{kparallel} as

\begin{equation}
k_{\text{parallel}}=2k_1+\frac{k_2}{2}.
\end{equation}

Therefore the spring array can we redrawn as in the previous figure.

\begin{figure}[H] \centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/array5.png}
\caption{Array after reduction}
\label{fig:array5}
\end{figure}

Finally, the resulting springs are in series, thus we can use equation \eqref{kseries1} to reduce the system to one spring of equivalent constant given by

\begin{equation}
\label{equiv}
k_{\text{series}}=\left(\frac{1}{2k_1+\frac{k_2}{2}}+\frac{1}{2k_3}\right)^{-1}.
\end{equation}

We can draw the system totally reduced as seen in the figure below.

\begin{figure}[H] \centering
\includegraphics[scale=0.6]{Nico y Jose/Img/Ondas/array6.png}
\caption{Array after reduction}
\label{fig:array6}
\end{figure}

For a mass spring system with no friction the angular frequency of oscillation \(\omega\) is given by

\begin{equation}
\label{omega}
\omega=\sqrt{\frac{k}{m}},
\end{equation}

where \(k\) is the equivalent spring constant and \(m\) the mass attached to the springs. Since we are asked about the frequency of oscillation \(f\), we can use the relation between \(f\) and \(\omega\) that reads

\begin{equation}
\label{efe}
f=\frac{\omega}{2\pi}.
\end{equation}

Using the expression given for \(\omega\) in equation \eqref{omega} into equation \eqref{efe} we get

\begin{equation}
f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}.
\end{equation}

Now, if we put our result for the equivalent constant \(k\) given in equation \eqref{equiv} into the equation above ge obtain

\begin{equation}
f=\frac{1}{2\pi}\sqrt{\frac{\left(\frac{1}{2k_1+\frac{k_2}{2}}+\frac{1}{2k_3}\right)^{-1}}{m}}.
\end{equation}

Using the numerical values for all variables we get

\begin{equation}
f=\frac{1}{2\pi}\sqrt{\frac{\left(\frac{1}{2(30\,\text{N/m})+\frac{50\,\text{N/m}}{2}}+\frac{1}{2(70\,\text{N/m})}\right)^{-1}}{(4\,\text{kg})}},
\end{equation}

\begin{equation}
f\approx 0.58\,\text{Hz}.
\end{equation}