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Sagot :
a)
In order to calculate the frequency of oscillations of a simple harmonic oscillator, first let's calculate the period, using the formula:
[tex]\begin{gathered} T=2\pi\sqrt[]{\frac{m}{k}} \\ T=2\pi\sqrt[]{\frac{2}{800}} \\ T=2\pi\sqrt[]{\frac{1}{400}} \\ T=2\pi\cdot\frac{1}{20} \\ T=\frac{\pi}{10}=0.31416\text{ s} \end{gathered}[/tex]Since the frequency is the inverse of the period, we have:
[tex]\begin{gathered} f=\frac{1}{T} \\ f=\frac{1}{\frac{\pi}{10}} \\ f=\frac{10}{\pi}=3.183\text{ Hz} \end{gathered}[/tex]b)
In order to calculate the maximum oscillation amplitude given the maximum speed, let's use the formula below:
[tex]\begin{gathered} v_{\max }=X\sqrt[]{\frac{k}{m}} \\ 10=X\cdot\sqrt[]{\frac{800}{2}} \\ 10=X\cdot20 \\ X=\frac{10}{20}=0.5\text{ meters} \end{gathered}[/tex]c)
The formula for the pendulum frequency is:
[tex]\begin{gathered} f=\frac{1}{T} \\ f=\frac{1}{2\pi\sqrt[]{\frac{L}{g}}} \\ f=\frac{\sqrt[]{\frac{g}{L}}}{2\pi} \end{gathered}[/tex]Then, for f = 3.183 and g = 9.81, we have:
[tex]\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}=\frac{3.132}{20} \\ \sqrt[]{L}=0.1566 \\ L=0.3957 \end{gathered}[/tex]So the length is 0.3957 m or 39.57 cm.
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