Answer: $2 m$

Solution. The period of oscillations of a mathematical pendulum of length $l$ is determined by the expression $T_{1}=2 \pi \sqrt{\frac{l}{g}}$.

And for a pendulum of length $\frac{l}{2}$, the period of oscillations is $T_{2}=2 \pi \sqrt{\frac{l}{2 g}}$.

From the condition of the problem, it follows that $T=\frac{T_{1}+T_{2}}{2}$.

As a result, we get $l=\frac{2 T^{2} g}{\pi^{2}(3+2 \sqrt{2})}=2 \mathrm{m}$.
6. (15 points) A mathematical pendulum is suspended near a vertical wall and oscillates in a plane parallel to the wall. A thin nail is driven into the wall such that the midpoint of the pendulum string hits it every time the pendulum passes through the equilibrium position from right to left. Find the length of the string if the period of oscillations of such a pendulum (with the disturbance caused by the nail) \( T = 2.41 \, \text{s} \). The acceleration due to gravity \( g = 10 \, \text{m/s}^2 \).

![](https://cdn.mathpix.com/cropped/2024_05_06_9c0cded10981f14efe49g-04.jpg?height=500&width=514&top_left_y=138&top_left_x=797)