Question

Two identical springs are connected in series and parallel as shown in the figure. If $ f_{s} $ and $ f_{p} $ are frequency of series and parallel arrangement 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 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 given by
$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 $
$ \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_1k_2}{k_1+k_2} $
Hence, frequency of this arrangement is given by
$ f_s =\frac{1}{2\pi} \sqrt{\frac{k_1k_2}{(k_1+k_2)}} $
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}} =\frac{1}{2} $