Problem — From amplitude to k

Problem

A simple harmonic oscillator has total energy $E = 0{,}80\;\text{J}$, amplitude $A = 8{,}0\;\text{cm}$ and oscillates with period $T = 0{,}40\;\text{s}$. Calculate: (a) spring constant $k$; (b) mass $m$.

Solution

Data: $E = 0{,}80\;\text{J}$, $A = 0{,}08\;\text{m}$, $T = 0{,}40\;\text{s}$.

(a) The total energy is entirely elastic at the turning points: $E = \tfrac{1}{2}kA^2$, so $k = \frac{2E}{A^2} = \frac{2 \cdot 0{,}80}{0{,}08^2} = \frac{1{,}6}{0{,}0064} \quad\Rightarrow\quad \ev{k = 250\;\text{N/m}}$

(b) From the angular frequency $\omega = 2\pi/T = 2\pi/0{,}40 \approx 15{,}71\;\text{rad/s}$ and from $\omega = \sqrt{k/m}$: $m = \frac{k}{\omega^2} = \frac{250}{246{,}7} \quad\Rightarrow\quad \ev{m \approx 1{,}01\;\text{kg}}$

Links

Topics: Oscillazioni e moto armonico Concepts: Moto armonico semplice · Energia potenziale elastica · Conservazione dell’energia meccanica Objects: Molla