Explanation

(a) 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
$f_{p} = \frac{1}{2\pi}\ \sqrt{\frac{k_{p}}{m}}$
or $f_{p} = \frac{1}{2\pi}\ \sqrt{\frac{k_{1} + k_{2}}{m}}$
but $k_{1} = k_{2} = k$
$\therefore$ $f_{p} = \frac{1}{2\pi}\ \sqrt{\frac{2k}{m}}$ ..(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
$f_{s} = \frac{1}{2\pi}\ \sqrt{\frac{k_{1}k_{2}}{(k_{1} + k_{2})m}}$
or $f_{s} = \frac{1}{2\pi}\ \sqrt{\frac{k}{2m}}$ …(ii)
Dividing Eq. (ii) by Eq. (i), we get
$\frac{f_{s}}{f_{p}} = \frac{\frac{1}{2\pi}\ \sqrt{\frac{k}{2m}}}{\frac{1}{2\pi}\ \sqrt{\frac{2k}{m}}} = \sqrt{\frac{1}{4}}$
or $\frac{f_{s}}{f_{p}} = \frac{1}{2}$