Answer: We essentially have to find the variable L, provided the value of the duration of one complete swing, or the period:
[tex]T=2\pi\sqrt{\frac{L}{980}}\Rightarrow(1)[/tex]
The value of the variable T is a known quantity in the equation (1), therefore plugging it into the equation (1) gives the following equation:
[tex](1.5)=2\pi\sqrt{\frac{L}{980}}\Rightarrow(2)[/tex]
The solution is found by solving for the variable L in the equation (2), the steps for the answer are as follows:
[tex]\begin{gathered} \begin{equation*} (1.5)=2\pi\sqrt{\frac{L}{980}} \end{equation*} \\ \\ \\ (\frac{3}{2})=2\pi\sqrt{\frac{L}{980}} \\ \\ \frac{1}{2\pi}\times\frac{3}{2}=\sqrt{\frac{L}{980}} \\ \\ \\ (\frac{1}{2\pi}\times\frac{3}{2})^2=\frac{L}{980}\rightarrow(\frac{1}{4\pi^2}\times\frac{9}{4})=\frac{L}{980}\Rightarrow\frac{1}{\pi^2}\times\frac{9}{16}=\frac{L}{980} \\ \\ \\ L=(\frac{980\times9}{16})\times\frac{1}{\pi^2}=\frac{2205}{4}\times\frac{1}{\pi^2} \\ \\ \\ L=55.853302482 \\ \\ \\ L\approx55.9cm \end{gathered}[/tex]
The answer is 55.9 centimeters or Option C.
