A spring of constant $k = 100$ N/m carries a mass of $1$ kg. Its period of oscillation is:
A$\pi$ s, ignoring the square root in the formula
B$0.628$ s, since $T = 2\pi\sqrt{m/k} = 2\pi/10$
C$10$ s, the inverse of the angular frequency
D$1$ s, equal to $\sqrt{m/k}$ without the factor
Answer & Solution
Correct answer: B. $0.628$ s, since $T = 2\pi\sqrt{m/k} = 2\pi/10$
$T = 2\pi\sqrt{1/100} = 2\pi/10 \approx 0.628$ s.
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