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We have, T=2\pi\sqrt{\frac{m}{k}}, where
T\rightarrow Time-period,
m\rightarrow mass attached to the spring,
k\rightarrow spring-constant.
Now if spring is cut into half its spring-constant, k, doubles.
Hence time period, T, is decresed by \sqrt{2} times.
The above explanation is valid only if spring is ideal, i.e. massless.
A more general formula is T=2\pi \sqrt{\frac{m+\frac{m_{o}}{3}}{k}},
here, m_{o} \rightarrow mass of spring.
I believe you can find the modified time-period.
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