Can someone please derive $T=2\pi\sqrt{l/g}$ or prove it without using calculus? I don't know much calculus, but I want to know that how one derives the formula to find time period $T$ of a simple pendulum.
 A: This answer elaborates on the one given by JKL. The only reason to enter it in a separate answer is that it won't fit within the 500 character limit of a comment.

The pendulum bob can swing in any direction in the horizontal plane. Take the following motion pattern: the bob swings around in a horizontal circle (so that the pendulum wire sweeps out a vertical cone). With such a circular motion pattern the distance of the bob to the equilibrium point remains the same. 
Interestingly, this circular motion can be thought of as a combination of two harmonic oscillations, perpendicular to each other.
This is a general rule when descibing motion: If the motion is in two dimensions then you can decompose everything in two perpendicular components, and the laws of motion apply perfectly fine for the individual motion components. 
When the pendulum is not swinging it's like a plumb line. I will call the point where the bob then hangs the 'equilibrium point'.
In the following expressions I use the letter 'r' for distance of the bob to the equilibrium point.
For small deviation from the equilibrium point the force(-component) on the bob towards the equilibrium point is proportional to the angle of the pendulum wire. To a good approximation the following is valid for the force:
$F = mg\frac{r}{L}$
The factor r/L is there because the ratio of r and L is a measure of the angle.
As mentioned by JKL, for circular motion the required centripetal force is given by:
$F_c = m\omega^2r$
Equating the two:
$mg\frac{r}{L} = m\omega^2r$
m and r drop out:
$\frac{g}{L} = \omega^2$
A: You can get a proportionality using one of the most basic techniques in science: dimensional analysis. What you do is look at the dimensions of the various quantities in the problem and try to see how they can fit together. Here you want the period, which is a time. I'll write $[T]$ to indicate that it has dimensions of time. What could it depend on? If we're ignoring friction (which is often reasonable) the only things it could depend on are:


*

*The mass of the pendulum $m$, with dimensions of mass $[M]$

*The length of the pendulum $\ell$, with dimensions of length $[L]$

*The acceleration of gravity $g$, with dimensions $[L/T^2]$


You want to put this together in a way that gives you a time. The most general thing you can write down is
$$ m^a \ell^b g^c, $$
for some numbers $a,b,c$. The dimensions of this are
$$ [M^a L^{b+c} T^{-2c}], $$
which you can see by putting in the dimensions and multiplying it out. Equating the dimension to $[T^1]$ we find:
$$ \begin{array}{lcl}
a &=& 0, \\
b+c &=& 0, \\
-2c &=& 1.
\end{array}$$
From which you find $a=0,\ b=-c=1/2$. So we know that
$$ T = k \sqrt{\frac{\ell}{g}}, $$
where $k$ is some pure number (no dimensions) which we can't work out this way. It turns out that $k=2\pi$ but dimensional analysis can only take you so far. I don't know of a way to get $k$ without using at least a little bit of calculus. :)
Note: More generally the period can and does depend on the magnitude of the swing, given by $\theta_{max}$, the largest angle of deflection from the vertical. Angles are dimensionless quantities, so really we can only write
$$ T = \sqrt{\frac{\ell}{g}} f(\theta_\textrm{max}), $$
where $f(\theta_\textrm{max})$ is some function we need to work out using Newton's laws and calculus, but it turns out that for reasonably small swings $f(\theta_\textrm{max})\approx 2\pi$ independent of $\theta_\textrm{max}$.
A: SIMPLE PHYSICS FOR THE PEDULUM EQUATION: No Calculus, only some algebra.
Forgive me if I assume you only know some good algebra and some basic physics, such as the equations of circular motion and SHM (simple harmonic motion.) With these constraints in mind I will try to show you how to prove the equation you are working on, based only on pure physics arguments, some simplifying assumptions, and very little mathematics. Once you get done with the explanations, it will then look easy!
But first, a couple of things from circular motion and SHM: 
The speed of a particle in uniform circular motion around a circle of radius r is 
$v=\omega r$ 
where $\omega=2\pi/T$. The acceleration of the particle is
$a=\omega^2r$
Now, let us see how we can apply all these in the case of the pendulum. 
It is very crucial to understand that the equation you are trying to prove holds only for very small deviations of the bob (the mass) from its “middle” equilibrium position. 
Now imagine you have the mass $m$ hanging vertically from some fixed point, by a piece of light and inextensible string of length $L$. Give the mass a very small deviation from the equilibrium position. As the mass oscillates it is moving on an arc of a vertical circle of radius $r=L$, but the deviation must be small. The smaller the deviation the better, because otherwise even  using calculus will not lead to the simple equation we want to prove. In fact, although the formal mathematical derivation requires a small deviation, it does not set a limit how small it must be.  So we shall take it to be so small, that the centripetal force that holds the particle in the circular motion is the weight, $mg$, of the bob. That means the centripetal acceleration is $g$. Therefore, according to the second of the above equations we can write
$g=\omega^2L$
$g=\frac{4\pi^2}{T^2}L$
Solve the last equation for T to you get: $T=2\pi\sqrt{\frac{L}{g}}$. 
