To solve this question, we have to relate period (seconds to make a cycle) and its length.
We can relate them as:
[tex]T=2\pi\sqrt[]{\frac{L}{g}}[/tex]
where g is the acceleration due to gravity and L the length of the pendulum.
If T1=2.00 and T2=1.99, we can relate them as:
[tex]\begin{gathered} \frac{T_2}{T_1}=\sqrt[]{\frac{L_2}{L_1}} \\ \frac{L_2}{L_1}=(\frac{T_2}{T_1})^2=(\frac{1.99}{2.00})^2=0.995^2=0.990025 \\ L_1=\frac{L_2}{0.990025}\approx1.01L_2 \end{gathered}[/tex]
Then, we know that the length of should be 1% larger than it actually is.
As we do not know the actual length, we will use the first equation to calculate the actual length first and then the correct length for a period of 2 seconds.
[tex]\begin{gathered} T=2\pi\sqrt[]{\frac{L}{g}} \\ \frac{T}{2\pi}=\sqrt[]{\frac{L}{g}} \\ L=g(\frac{T}{2\pi})^2 \\ L=9.81\cdot(\frac{1.99}{2\cdot3.14})^2=9.81\cdot0.3167^2=9.81\cdot0.1=0.981\text{ m} \end{gathered}[/tex]
NOTE: all the variables and constants are in meters and seconds.
As the correct length is 1% larger than 0.981 m, we can calculate the increase in length as:
[tex]\Delta L=0.01\cdot L_2=0.01\cdot0.981m=0.00981\text{ m}[/tex]
Answer: 0.00981 m
