A block of mass 4.0kg causes a spiral spring to extend by 0.16m from its unstretched position. The block is removed and another body of mass 0.50kg is hung from the same spiral spring. If the spring is then stretched and released, what is the angular frequency of the subsequent motion? [g = 10 ms ^-2]

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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