Normal Mode in a Vibration? What exactly is the Normal Mode?  According to me, it forms a basis functions (where Max transverse amplitude is fixed wrt position), and all arbitrary vibration in the string can be written in the form of linear form of this basis function.
Please tell me where I am not grasping the concept?
 A: Yes, the trigonometric functions of the Fourier series form an orthogonal basis for the solution.
The differential equation for transversal waves in a string can be derived by applying the second Newton's law in an elementary length, which results in$$\frac{\partial^2 y}{\partial t^2} = \frac{T}{\rho}\frac{\partial^2 y}{\partial x^2}$$Without additional information about boundary conditions, any function $y(x,t)$ that can be made a sum of the form:$y(x,t) = f(x-vt) + g(x+vt)$, where $v = \sqrt{\frac{T}{\rho}}$, and $f$ and $g$ are arbitrary functions, is a solution.
When both ends of the string are supposed fixed, $y(x,t)$ can be imagined as a periodic function on $(-\infty,+\infty)$, where the length of the string is a period. Any function like that can be expressed as a Fourier series with infinite terms. For $t = 0$, $$y = \sum_{n=-\infty}^{+\infty}c_n\exp{\left(ik_n x\right)}$$ where $k_n = \frac{2\pi n}{L}$, $L$ is the length and $c_n$ are complex coeficients (so that the terms are real).
For $t\neq 0$, we can make the coeficients $c_n$ functions of the time: $c_n = A_n \exp({ik_nvt})$ or $c_n = A_n \exp({-ik_nvt})$. That means, for each $t$ there is a different function, expressed by a different Fourier series. This procedure works because, if we join the exponetials together:
$$y(x,t) = \sum_{n=-\infty}^{+\infty}A_n(\exp{ik_n(x+vt)} + \exp{ik_n(x-vt)})$$
that is a sum of functions of $x-vt$ and $x+vt$ as required for the solution.
Each $k_n$ defines a normal mode with an angular frequency $$\omega = k_n v = \frac{2\pi n}{L}\sqrt{\frac{T}{\rho}}$$
