Two identical springs are connected in series and parallel as shown in the figure. If $f_{s} a n d f_{p}$ are frequencies of arrangements, what is $\frac{f_{s}}{f_{p}}$ ?

Question

Two identical springs are connected in series and parallel as shown in the figure. If f s a n d f p are frequencies of arrangements, what is f s f p ?

TestHub · Accepted Answer

In first case, springs are connected in parallel, so their equivalent spring constant

$k_{p} = k_{1} + k_{2}$

So, frequency of this spring-block system is

$f_{p} = \frac{1}{2 π} \sqrt{\frac{k_{p}}{m}}$

or $f_{p} = \frac{1}{2 π} \sqrt{\frac{k_{1} + k_{2}}{m}}$

but $k_{1} = k_{2} = k$

$∴$ $f_{p} = \frac{1}{2 π} \sqrt{\frac{2 k}{m}}$ ..(i)

Now in second case, springs are connected in series, so their equivalent spring constant

$k = \frac{k_{1} k_{2}}{k_{1} + k_{2}}$

Hence, frequency of this arrangement is given by

$f_{s} = \frac{1}{2 π} \sqrt{\frac{k_{1} k_{2}}{(k_{1} + k_{2}) m}}$

or $f_{s} = \frac{1}{2 π} \sqrt{\frac{k}{2 m}}$ ...(ii)

Dividing Eq. (ii) by Eq. (i), we get

$\frac{f_{s}}{f_{p}} = \frac{\frac{1}{2 π} \sqrt{\frac{k}{2 m}}}{\frac{1}{2 π} \sqrt{\frac{2 k}{m}}} = \sqrt{\frac{1}{4}}$

or $\frac{f_{s}}{f_{p}} = \frac{1}{2}$

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