he equation of motion of a particle is𝑑2𝑦𝑑𝑡2 + 𝐾𝑦 = 0, where K is positiveconstant. The time period of the motionis given by

1. Break Down the Problem

We are given the second-order differential equation of motion: $\frac{d^2y}{dt^2} + Ky = 0$ Our goal is to find the time period of the motion.

2. Relevant Concepts

This is a simple harmonic motion (SHM) equation, where the general form is: $\frac{d^2y}{dt^2} + \omega^2 y = 0$ where $\omega^2 = K$. The solution to this equation is: $y(t) = A \cos(\omega t + \phi)$ where $A$ is the amplitude and $\phi$ is the phase constant.

From the properties of SHM, the time period $T$ can be expressed as: $T = \frac{2\pi}{\omega}$

3. Analysis and Detail

Since we identified that: $\omega^2 = K \implies \omega = \sqrt{K}$ We can substitute $\omega$ into the formula for time period $T$: $T = \frac{2\pi}{\sqrt{K}}$

4. Verify and Summarize

To verify, we recall the original equation and the relationship with time period derived from the properties of SHM. Both the equation and derived expressions check out mathematically.

Final Answer

The time period of the motion is: $T = \frac{2\pi}{\sqrt{K}}$