# Springs in series and in parallel

## Homework Statement

A mass ##m## hangs from a combination of two springs, each with spring constant ##k##, connected in series. If the mass is doubled to ##2m## the mass will hang lower by a distance ##h##. If three such springs are arranged in parallel to support a mass of ##5m## what will be the frequency of small oscillations if the system is perturbed?

## The Attempt at a Solution

A combination of two springs in series, each with spring constant ##k##, is equivalent to a system of one spring with spring constant
\begin{align}
\frac{1}{k_{\text{eq}}} &= \frac{1}{k} + \frac{1}{k}\\
k_{\text{eq}} &= k/2.
\end{align}
For a mass ##m## hanging from the system of two springs, let the extension of the springs be ##x## m. Then, if the mass is doubled to ##2m##, the extension of the springs is ##(x + h)## m. Therefore, we find that
\begin{align}
mg &= (k/2)x,\\
2mg &= (k/2)(x + h),
\end{align}
so that
\begin{align}
\frac{x + h}{x} &= 2\\
x &= h.
\end{align}
Therefore, the spring constant of each spring, in terms of the mass ##m## and distance ##h##, is
\begin{align}
k = 2mg/h.
\end{align}
A combination of three springs in parallel, each with spring constant ##2mg/h##, is equivalent to a system of one spring with spring constant
\begin{align}
k_{\text{eq}} &= 2mg/h + 2mg/h + 2mg/h\\
k_{\text{eq}} &= 6mg/h.
\end{align}
If the springs are arranged in parallel to support a mass of ##5m## and perturbed from equilibrium, then the frequency of small oscillations is
\begin{align}
\omega = \sqrt{k_{\text{eq}}/m} = \sqrt{6g/h}.
\end{align}

Is my solution correct?

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Doc Al
Mentor
If the springs are arranged in parallel to support a mass of ##5m## and perturbed from equilibrium, then the frequency of small oscillations is
\begin{align}
\omega = \sqrt{k_{\text{eq}}/m} = \sqrt{6g/h}.
\end{align}
The mass is 5m, not m. (Otherwise good.)

• Afonso Campos
Ah, right! So, the answer is

$$\omega = \sqrt{k_{\text{eq}}/5m} = \sqrt{6g/5m}.$$

Doc Al
Mentor
Looks good.