A spring stretches by $2$ mm when it is loaded with a mass of $200$ g. From equilibrium position the mass is further pulled down by $2$ mm and released. The frequency associated with the system and maxmimum energy in the spring are ______ Hz and ______ J, respectively. (Take $g = 10\ \text{m/s}^2$)
Answer: (A) $\dfrac{5\sqrt{50}}{\pi}$ and $8 \times 10^{-3}$
Concept. The static stretch gives $k=\dfrac{mg}{x_0}$, and $\omega=\sqrt{k/m}=\sqrt{g/x_0}$. The spring's energy is greatest at the lowest point, where its total stretch is $x_0+A$.
Step 1. Spring constant. $k=\dfrac{0.2\times10}{0.002}=1000$ N m$^{-1}$.
Step 2. Frequency. $\omega=\sqrt{\dfrac{1000}{0.2}}=\sqrt{5000}=10\sqrt{50}$; $f=\dfrac{\omega}{2\pi}=\dfrac{5\sqrt{50}}{\pi}$ Hz.
Step 3. Maximum spring energy. Stretch $2+2=4$ mm: $\tfrac12\times1000\times(0.004)^2=8\times10^{-3}$ J.
Answer. Option A. Trap: the SHM energy $\tfrac12kA^2=2\times10^{-3}$ J is not the spring's stored energy.
Solution by Sreeraj P, M.Sc Physics
