# Springs in series and in parallel

1. Jun 11, 2017

### Afonso Campos

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

3. 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?

2. Jun 11, 2017

### Staff: Mentor

The mass is 5m, not m. (Otherwise good.)

3. Jun 11, 2017

### Afonso Campos

Ah, right! So, the answer is

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

4. Jun 11, 2017

Looks good.