Question

Two identical springs of spring constant $k$ are connected in series and paralIeI as shown in figure. $A$ mass $M$ is suspended from them. The ratio of their frequencies of vertical oscillation will be :

• A. $1:2$
• B. $2:1$
• C. $4:1$
• D. $1:4$

Answer 1

The correct option is A $1:2$

We know that: $T=2\pi \sqrt{\frac{m}{K}}\Rightarrow n=\frac{1}{2\pi }\sqrt{\frac{K}{m}}$
For a spring mass system.
In case 1: If $K$ is the resultant spring constant, then$\frac{1}{K}=\frac{1}{k}+\frac{1}{k}=\frac{2}{k}\Rightarrow K=\frac{k}{2}$
In case 2, $K=k+k=2k$

If $n_1$ & $n_2$ be frequencies in two cases, then
$n_1=\frac{1}{2\pi }\sqrt{\frac{k}{2m}};n_2=\frac{1}{2\pi }\sqrt{\frac{2k}{m}};$
$\Rightarrow \frac{n_1}{n_2}=\sqrt{\frac{1}{4}}\Rightarrow \frac{n_1}{n_2}=\frac{1}{2}$