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two different spring systems

Two identical springs with spring constant $k$ are connected to identical masses of mass $M$, as shown in the figures above. The ratio of the period for the springs connected in parallel (Figure 1) to the period for the springs connected in the series (Figure 2) is $ 1/2 $

What would be the better way to solve this? I have used this law $$\begin{equation} T = 2 \pi \sqrt{\frac{l}{g}} \end{equation}$$ and assumed, $2l$ for the $2^{nd}$ picture but got wrong answer.

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up vote 2 down vote accepted

Two springs in parallel effectively behave as a single spring with spring constant $k_\mathrm{parallel} = k_1+k_2 = 2k$ while two springs in series effectively behave as a single spring with spring constant $k_\mathrm{series} = k_1k_2/(k_1+k_2) = k/2$. Recall that the period of a mass on a spring is $$ T = 2\pi \sqrt{\frac{m}{k}} $$ Which gives $$ \frac{T_\mathrm{parallel}}{T_\mathrm{series}} = \sqrt{\frac{k_\mathrm{series}}{k_\mathrm{parallel}}} = \sqrt{\frac{k/2}{2k}} = \frac{1}{2} $$

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can you prove them (the series and parallel formulas) too? – ABC May 9 '13 at 7:22
@007 The proofs are here – joshphysics May 9 '13 at 7:24
$k_\mathrm{series} = k_1k_2/(k_1+k_2) = k/2$ how this can be written? – Sonia Afrin May 9 '13 at 7:33
In this case, $k_1 = k_2 = k$. – joshphysics May 9 '13 at 7:36
Got it :-) thnx – Sonia Afrin May 9 '13 at 7:38

You are using the wrong law .

This is linear SHM. Find out the time period for that , it will be $2\pi \sqrt \frac{m}{k}$


Then figure 1 , springs are in parallel and in figure 2 , springs are in series . Do a wiki search to figure out how to find equivalent springs .

An Advice : Don't learn equations blindly .

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nonagon, @007: Please keep it civil. Both of you. We expect people to be nice on this site. Take this as a warning; further incidents may lead to a suspension. – Manishearth May 9 '13 at 12:16
007 is correct in saying that it would be best if you included formulae. However, there is no compulsion to do so. The answer is OK as is. Insulting each other's intelligence is not a way to carry out a constructive discussion; please avoid personal attacks. – Manishearth May 9 '13 at 12:17

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