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## 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?

## Homework Equations

## 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?