Question

Two identical springs of constant are connected in series and parallel as shown in figure. A mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be

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

Answer 1

Answer: (C)

$1:2$

We know that, $n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

In series, $k=\frac{k_1{k}_2}{k_1+k_2}=\frac{k^2}{2}$

In parallel, $k=k_1+k_2=2k$

$\therefore ,n_1=\frac{1}{2\pi }\sqrt{\frac{k}{2m}}$

and $n_2=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}$

hence, $n_1:n_2=1:2$