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. 1:02
• B. 2:01
• C. 1:03
• D. 3:01

Answer 1

Answer: (A)

1:02

Answer 2

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 E (ii) by E (i), we get $\frac{f_s}{f_p} =\frac{\frac{1}{2\pi} \sqrt{\frac{k}{2m}}}{\frac{1}{2\pi} \sqrt{\frac{2k}{m}}}$ $=\sqrt{\frac{1}{4}} =\frac{1}{2}$