Question

when a spring is cut into half its time period should remain same because mass halves and k doubles right ?

Answer 1

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.