Problem — Simple harmonic oscillator (250 g)

Problem

A mass of $250\;\text{g}$ attached to a horizontal spring with $k = 40\;\text{N/m}$, displaced $5\;\text{cm}$ from the equilibrium position and released. Calculate: (a) angular frequency, (b) period, (c) maximum velocity, (d) total energy.

Solution

Data: $m = 0{,}25\;\text{kg}$, $k = 40\;\text{N/m}$, $A = 0{,}05\;\text{m}$.

(a) Angular frequency: $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{40}{0{,}25}} = \sqrt{160} \quad\Rightarrow\quad \ev{\omega \approx 12{,}65\;\text{rad/s}}$

(b) Period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{12{,}65} \quad\Rightarrow\quad \ev{T \approx 0{,}497\;\text{s}}$

(c) Maximum velocity (at the passage through equilibrium): $v_{\max} = A\,\omega = 0{,}05 \cdot 12{,}65 \quad\Rightarrow\quad \ev{v_{\max} \approx 0{,}633\;\text{m/s}}$

(d) Total energy: $E = \tfrac{1}{2}kA^2 = \tfrac{1}{2}\cdot 40 \cdot 0{,}05^2 \quad\Rightarrow\quad \ev{E = 0{,}05\;\text{J}}$

Links

Topics: Oscillazioni e moto armonico Concepts: Moto armonico semplice · Periodo e frequenza · Conservazione dell’energia meccanica Objects: Molla