ANSWER:
0.01625 N
STEP-BY-STEP EXPLANATION:
Given:
mass = m = 13 g = 0.013 kg
period = p = 0.8 sec
length = L = 20 cm = 0.2 m
We have the following formula to calculate the tension in the string:
[tex]p=\frac{2L}{\sqrt[]{\frac{T}{m\text{/L}}}}[/tex]We solve for T:
[tex]\begin{gathered} p^2=\frac{4L^2}{\frac{T}{m\text{/L}}} \\ p^2=\frac{4L^2\cdot m}{TL} \\ T=\frac{4L\cdot m}{p^2} \\ \text{ We replace} \\ T=\frac{4\cdot0.2\cdot0.013}{0.8^2} \\ T=0.01625\text{ N} \end{gathered}[/tex]The tension in the string is 0.01625 N
