15-26.

This one is a bit tricky since it's not obvious how to apply the results we have for spring systems to this case. There are a few different ways to look at this, but consider the following. Both springs have the same spring constant, $k$. If we move the mass to the left, symmetry demand than both springs stretch by an equal amount. So if the distance of the mass from equilibrium is $x$, and of the two springs $x_1$ and $x_2$, then $x_1=x_2=x/2$. Now, the potential energy of the two spring system is given by:

$$\begin{eqnarray*} U&=&\frac{1}{2}kx_1^2+\frac{1}{2}kx_2^2\\ &=&\frac{k}{2}\left(x_1^2+x_2^2\right)\\ &=&k(x/2)^2\\ &=& kx^2/4. \end{eqnarray*}$$

Thus the force is

$$F=-\frac{dU}{dx}=-\frac{1}{2}kx,$$

So that the effective spring constant is cut in half. Thus the new frequency of oscillations will be given by

$$f'=\frac{1}{2\pi}\sqrt{\frac{k}{2m}}=\frac{f}{\sqrt{2}}=18.23\;\textrm{Hz},$$

where $f$ would be the frequency if only one spring were present.

This result should make sense: if a spring is made longer, then a given stretch of that spring is spead out over the the spring more. Thus there is less stress on each part of the spring and less force. Thus the overall spring constant is decreased. From the above formula we see that this descrease is linear, that is

$$k\propto \frac{1}{L}$$

for a given material composition, where $L$ is the spring length.