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

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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