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

Solution

## The correct option is A $$1 : 2$$We know that: $$\displaystyle T=2\pi \sqrt {\frac {m}{K}} \Rightarrow n= \frac {1}{2\pi} \sqrt {\frac {K}{m}}$$For a spring mass system.In case 1: If $$K$$ is the resultant spring constant, then$$\displaystyle \frac {1}{K} = \frac {1}{k} + \frac {1}{k} = \frac {2}{k} \Rightarrow K = \frac {k}{2}$$In case 2, $$K = k + k =2k$$If $$n_1$$ & $$n_2$$ be frequencies in two cases, then$$\displaystyle n_1 = \frac {1}{2\pi} \sqrt {\frac {k}{2m}}; n_2 = \frac {1}{2\pi} \sqrt {\frac {2k}{m}};$$$$\displaystyle \Rightarrow \frac {n_1}{n_2} = \sqrt {\frac {1}{4}} \Rightarrow \frac {n_1}{n_2} = \frac {1}{2}$$Physics

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