Question

Two identical springs are connected in series and parallel as shown in the figure. If $f_{s} \, and \, f_{p}$ are frequencies of 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 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 $ $f_{p}=\frac{1}{2 \pi } \, \sqrt{\frac{2 k}{m}}$ ..(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
$f_{s}=\frac{1}{2 \pi } \, \sqrt{\frac{k_{1} k_{2}}{\left(\right. k_{1} + k_{2} \left.\right) m}}$
or $f_{s}=\frac{1}{2 \pi } \, \sqrt{\frac{k}{2 m}}$ ...(ii)
Dividing Eq. (ii) by Eq. (i), we get
$\frac{f_{s}}{f_{p}}=\frac{\frac{1}{2 \pi } \, \sqrt{\frac{k}{2 m}}}{\frac{1}{2 \pi } \, \sqrt{\frac{2 k}{m}}}=\sqrt{\frac{1}{4}}$
or $\frac{f_{s}}{f_{p}}=\frac{1}{2}$