Two identical spring of constant K are connected in series and parallel as shown in figure. A mass m

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13.Oscillations

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Two identical spring of constant $K$ 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

$4:1$

Solution

(c) $n = \frac{1}{{2\pi }}\sqrt {\frac{k}{m}} $

==> $\frac{{{n_S}}}{{{n_P}}} = \sqrt {\frac{{{k_S}}}{{{k_P}}}} $

$ \Rightarrow \frac{{{n_s}}}{{{n_p}}} = \sqrt {\frac{{\left( {\frac{k}{2}} \right)}}{{2k}}} = \frac{1}{2}$

Standard 11

Physics

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