10\(\sqrt 5\) rads-1
5\(\sqrt 2\) rads-1
5 rads-1
\(\sqrt 5\) rads-1
Correct answer is A
K = \(\frac{F}{e} = \frac{4}{0.16} = 250\)
F = \(\frac{1}{T} = \frac{1}{2 \pi \frac{\sqrt{m}}{k}}\)
= \(\frac{1}{2 pi \frac{\sqrt{0.5}}{250}}\)
= 10\(\sqrt{5}\)
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