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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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