The frequency of oscillation of a mass $$m$$ suspended by a spring is $$v_1$$. If the length of the spring is cut to half, the same mass oscillates with frequency $$v_2$$. The value of $$\frac{v_2}{v_1}$$ is __________.

For a spring-mass system:

$$\nu=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

For a spring:

$$k\propto\frac{1}{L}$$

So if length is halved:

$$k_2=2k_1$$

Now frequency:

$$nu_2=\frac{1}{2\pi}\sqrt{\frac{2k_1}{m}}=\sqrt{2}$$

final answer:

$$\frac{\nu_2}{\nu_1}=\sqrt{2}$$

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