Answer :
you have to increase 0.01 meters the length
Explanation
The period of a pendulum is given by:
[tex]T=2\text{ }\pi\sqrt[]{\frac{L}{g}}[/tex]where T is the period, L is the length , a g is the the acceleration of the gravity (9.8 m per square second)
s
Step 1
find the length, for period= 2 s
Let
[tex]T_1=2\text{ seconds}[/tex]with this value, we can find the length
replace, and isolate L
[tex]\begin{gathered} T=2\text{ }\pi\sqrt[]{\frac{L}{g}} \\ 2=2\text{ }\pi\sqrt[]{\frac{L}{g}} \\ \text{divide both sides by 2}\pi \\ \frac{2}{2\pi}=\frac{2\text{ }\pi}{2\pi}\sqrt[]{\frac{L}{g}} \\ \frac{1}{\pi}=\sqrt[]{\frac{L}{g}} \\ (\frac{1}{\pi})^2=(\sqrt[]{\frac{L}{g}})^2 \\ \frac{1}{\pi^2}=\frac{L}{g} \\ \text{Multiply both sides by g} \\ \frac{1}{\pi^2}\cdot g\cdot g=\frac{L}{g} \\ \frac{g}{\pi^2}=L \end{gathered}[/tex]so, when the period is 2.00 s the length of the pendulum is
[tex]\begin{gathered} L_1=\frac{g}{\pi^2} \\ L_1=\frac{9.8}{\pi^2} \\ L_1=0.9929475997\text{ m} \end{gathered}[/tex]Step 2
now, for Period = 1.99 s
Let
[tex]T=1.99\text{ s}[/tex]replace and solve for L
[tex]\begin{gathered} T=2\text{ }\pi\sqrt[]{\frac{L}{g}} \\ 1.99=2\text{ }\pi\sqrt[]{\frac{L}{g}} \\ \text{divide both sides by 2}\pi \\ \frac{1.99}{2\pi}=\frac{2\text{ }\pi}{2\pi}\sqrt[]{\frac{L}{g}} \\ \frac{1.99}{2\pi}=\sqrt[]{\frac{L}{g}} \\ (\frac{1.99}{2\pi})^2=(\sqrt[]{\frac{L}{g}})^2 \\ \\ 0.1003105048=\frac{L}{g} \\ \text{Multiply both sides by g} \\ 0.1003105048\cdot g=\frac{L}{g}\cdot g \\ 0.1003105048\cdot g=L \\ L=0.1003105048\cdot9.8 \\ L_2=0.983049474\text{ m} \end{gathered}[/tex]so, when the period is 1.99 s , the length is 0.9830429474 m
Step 3
finally, to know how much you must increase the length, subtract L2 from L1
so
[tex]\begin{gathered} L_1-L_2=0.9929475997\text{ m-}0.9830429474m \\ L_1-L_2\approx0.99-0.98 \\ L_1-L_2\approx0.01 \end{gathered}[/tex]therefore, the make the pendulum osscilates with period of 2 seconds, you have to increase 0.01 meters the length
I hope this helps you