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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 :

430252_f585ce02f2164d7fae69043a83c9e470.png


A
1:2
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B
2:1
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C
4:1
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D
1:4
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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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