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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
632364_e5ef6afe5eaa4f61a25f0c5f1acfb9f3.PNG


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

The correct option is A $$1:2$$
We know that, $$n=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$$
In series, $$k=\dfrac{k_1k_2}{k_1+k_2}=\dfrac{k^2}{2}$$
In parallel, $$k=k_1+k_2=2k$$
$$\therefore, n_1=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{2m}}$$
and $$n_2 = \dfrac{1}{2\pi}\sqrt{\dfrac{2k}{m}}$$
hence, $$n_1:n_2=1:2$$

Physics

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