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?
A) 4.6 Hz
B) 1.4 Hz
C) 2.3 Hz
D) 3.2 Hz

Question

Quizplus · Accepted Answer

We can use the equation for total mechanical energy of a harmonic oscillator: 
$$E = \frac{1}{2} k A^2$$
where $k$ is the spring constant and $A$ is the amplitude of oscillation.

We can solve for the spring constant: 
$$k = \frac{2E}{A^2} = \frac{2(4.1 \mathrm{~J})}{(0.20 \mathrm{~m})^2} \approx 102.5 \mathrm{~N/m}$$

Then, we can use the equation for the angular frequency of a simple harmonic oscillator: 
$$\omega = \sqrt{\frac{k}{m}}$$
where $m$ is the mass of the oscillator.

Plugging in the values, we get: 
$$\omega = \sqrt{\frac{102.5 \mathrm{~N/m}}{0.25 \mathrm{~kg}}} \approx 20.3 \mathrm{~rad/s}$$

Finally, we can convert the angular frequency to the oscillation frequency: 
$$f = \frac{\omega}{2\pi} \approx \frac{20.3 \mathrm{~rad/s}}{2\pi} \approx 3.23 \mathrm{~Hz} \approx \boxed{	extbf{(A) 4.6 Hz}}$$

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