Answer
(a) $T' = 2.83~s$
(b) $T' = 1.41~s$
(c) $T' = 2.00~s$
(d) $T' = 1.41~s$
Work Step by Step
We can write an expression for the period $T$ of a mass oscillating on a spring;
$T = 2\pi~\sqrt{\frac{m}{k}} = 2.00~s$
(a) We can find the period $T'$ when the mass is doubled as:
$T' = 2\pi~\sqrt{\frac{2m}{k}}$
$T' = \sqrt{2}\times 2\pi~\sqrt{\frac{m}{k}}$
$T' = \sqrt{2}\times T$
$T' = (\sqrt{2})(2.00~s)$
$T' = 2.83~s$
(b) We can find the period $T'$ when the mass is halved as:
$T' = 2\pi~\sqrt{\frac{m/2}{k}}$
$T' = \sqrt{\frac{1}{2}}\times 2\pi~\sqrt{\frac{m}{k}}$
$T' = \sqrt{\frac{1}{2}}\times T$
$T' = (\sqrt{\frac{1}{2}})(2.00~s)$
$T' = 1.41~s$
(c) Since the period does not depend on the oscillation amplitude, the period of $T = 2.00~s$ remains the same.
(d) We can find the period $T'$ when the spring constant is doubled;
$T' = 2\pi~\sqrt{\frac{m}{2k}}$
$T' = \sqrt{\frac{1}{2}}\times 2\pi~\sqrt{\frac{m}{k}}$
$T' = \sqrt{\frac{1}{2}}\times T$
$T' = (\sqrt{\frac{1}{2}})(2.00~s)$
$T' = 1.41~s$