Question

Two identical springs are connected in series and parallel as shown in the figure. If $ f_s $ and $ f_p $ are frequencies of series and parallel arrangements, what is $ \frac{f_s}{f_P} $ ?

• A. $ 1 : 2 $
• B. $ 2 : 1 $
• C. $ 1 : 3 $
• D. $ 3 : 1 $

Answer 1

Answer: (A)

In first arrangement springs are in parallel and in second arrangement springs are in series.
In first case, springs are connected in parallel, so their equivqlent 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}}\quad...\left(i\right)$
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}}{\left(k_{1}+k_{2}\right)m}} $
or $f_{s} = \frac{1}{2\pi}\sqrt{\frac{k}{2m}}\quad...\left(ii\right) $
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}$