Question

A simple harmonic oscillator has a mass m=2.0 kg and a spring constant k=800 N/m.a) What is the frequency of oscillations?b) If the oscillator has a maximum speed of v=10.0 m/s, what is the maximum oscillation amplitude?c) What length would a simple pendulum need to be in order to oscillate at the same frequency?

Answer 1

a)

In order to calculate the frequency of oscillations of a simple harmonic oscillator, first let's calculate the period, using the formula:

`\begin{gathered} T=2\pi\sqrt[]{(m)/(k)} \\ T=2\pi\sqrt[]{(2)/(800)} \\ T=2\pi\sqrt[]{(1)/(400)} \\ T=2\pi\cdot(1)/(20) \\ T=(\pi)/(10)=0.31416\text{ s} \end{gathered}`

Since the frequency is the inverse of the period, we have:

`\begin{gathered} f=(1)/(T) \\ f=(1)/((\pi)/(10)) \\ f=(10)/(\pi)=3.183\text{ Hz} \end{gathered}`

b)

In order to calculate the maximum oscillation amplitude given the maximum speed, let's use the formula below:

`\begin{gathered} v_(\max )=X\sqrt[]{(k)/(m)} \\ 10=X\cdot\sqrt[]{(800)/(2)} \\ 10=X\cdot20 \\ X=(10)/(20)=0.5\text{ meters} \end{gathered}`

c)

The formula for the pendulum frequency is:

`\begin{gathered} f=(1)/(T) \\ f=\frac{1}{2\pi\sqrt[]{(L)/(g)}} \\ f=\frac{\sqrt[]{(g)/(L)}}{2\pi} \end{gathered}`

Then, for f = 3.183 and g = 9.81, we have:

`\begin{gathered} 3.183=\frac{\frac{\sqrt[]{9.81}}{\sqrt[]{L}}}{2\pi} \\ \frac{\sqrt[]{9.81}}{\sqrt[]{L}}=3.183\cdot2\pi \\ \frac{3.132}{\sqrt[]{L}}=20 \\ \sqrt[]{L}=(3.132)/(20) \\ \sqrt[]{L}=0.1566 \\ L=0.3957 \end{gathered}`

So the length is 0.3957 m or 39.57 cm.