A mass of \(0.5 kg\) is attached to one end of a helical spring and produces an extension of \(2.5 cm\). The mass now set into vertical oscillation of amplitude \(10 mm\). The period of oscillation is ____________ \(\left(g=10 m / s ^{2}\right)\)
A. \(0.33\) s B. \(100 s\) C. \(200 s\) D. \(280 s\)
Correct Answer: A
Explanation
$$ \omega=\sqrt{\frac{k}{m}} $$ since \(f=m g=k e \omega=\sqrt{\frac{m g / e}{m}}\); $$ \omega=\sqrt{\frac{g}{e}} $$lso, \(\omega=2 \pi f=\sqrt{\frac{g}{e}}\); $$ \begin{aligned} &f=\frac{1}{2 \pi} \sqrt{\frac{g}{e}} ; T=2 \pi \sqrt{\frac{e}{g}} \\ &t=2 \pi \sqrt{\frac{0.025}{10}}=0.314 \end{aligned} $$oat speed \(=\frac{\left(\begin{array}{l}\text { distance from } \\ \text { the } \\ \text { drop point }\end{array}\right)}{\text { time }}\) \(=\frac{s}{t}------(1)\) \(s=25 m , t=\) ?
