1. **Problem Statement:** We have a system where the $m^{th}$ mode of vibration is on a string with length $2m$, and the $2m^{th}$ mode is related to a point $P$ on the $P$-axis. There is a circular particle $A$ centered at $P$ with radius $x$. We want to find the distance $x$ from $P$ to the particle $A$ and the angular frequency $\omega = \sqrt{\frac{g}{a}}$. 2. **Given:** - Distance along $P$ axis is $a/2$ on either side. - $AP = x$. - $\omega = \sqrt{\frac{g}{a}}$. - $O$ and $P$ are centers of particles and circle. - $l$ is the distance traveled by the particle. 3. **Step 1: Express the position of particle $O$ relative to $P$** Since $O$ is at a distance $x$ from $P$ on the $P$ axis, and $P$ is at $a/2$ from the origin, the position of $O$ is: $$ O = P \pm x = \frac{a}{2} \pm x $$ 4. **Step 2: Relation between modes and position** The $m^{th}$ mode corresponds to a wavelength related to $2m$, and the $2m^{th}$ mode corresponds to $P$ axis. The particle's position $x$ relates to these modes. 5. **Step 3: Using the formula for angular frequency** Given: $$ \omega = \sqrt{\frac{g}{a}} $$ where $g$ is acceleration due to gravity and $a$ is a length parameter. 6. **Step 4: Final expression for $x$** From the problem setup and symmetry, the distance $x$ from $P$ to $A$ is: $$ x = \frac{a}{2} $$ 7. **Summary:** - The distance $x$ from $P$ to the particle $A$ is $\frac{a}{2}$. - The angular frequency $\omega$ is $\sqrt{\frac{g}{a}}$. Thus, the solution is: $$ x = \frac{a}{2}, \quad \omega = \sqrt{\frac{g}{a}} $$