A 0.25 kg harmonic oscillator has a total mechanical energy of $4.1 \mathrm {~J}$ If the oscillation amplitude is $20.0 \mathrm {~cm}$ what is the oscillation frequency?

Question

Quizplus · Accepted Answer

The total mechanical energy $E$ of a harmonic oscillator is given by $E = \frac{1}{2} k A^2$, where $k$ is the spring constant and $A$ is the amplitude. The angular frequency $\omega$ is related to the spring constant and mass $m$ by $\omega = \sqrt{\frac{k}{m}}$. The frequency $f$ is $\frac{\omega}{2\pi}$. Given $E = 4.1 \, \text{J}$ and $A = 0.2 \, \text{m}$, we can find $k$ from the energy equation: $4.1 = \frac{1}{2} k (0.2)^2$, solving for $k$ gives $k = 205 \, \text{N/m}$. Then, $\omega = \sqrt{\frac{205}{0.25}}$, and $f = \frac{\omega}{2\pi}$, which calculates to approximately $4.6 \, \text{Hz}$.

A 025 Kg Harmonic Oscillator Has a Total Mechanical Energy Of Of 4.1