S H M

you can always write the differential equation and solve it [3]

using the eqns of motion only for a pendulum with a mass of bob - m

and length L when released from an angle of θ₀

we can easily get the expression for time period as

T=4\sqrt{\frac{l}{2g}}\int_{0}^{\theta_0}{\frac{d\theta }{\sqrt{cos\theta -cos\theta _0}}}

for very small θ₀ this gives the known time period of a simple pendulum for small oscillations..

i dont know if you meant this..for this is quite a well known expression

The principle of least action is way out of syllabus... so maybe not now..

besides it provides just another route to the answer ...at times easier... the eqn we get after solvin the langrangian can usually be got using simple force equations

Absolutely marvellous quotes . Subho and all the others -- the exact solution is howver as follows --

Suppose the bob of the pendulum is displaced by an angle x as shown in the figure .

K . E . of the pendulum at the point A = 12 m ( l dxdt ) ² =12 m ( l x₁ )² , where x₁ stands for the

derivative of x w.r.t time .

P . E . = m g l ( 1 - cos x )

Hence the lagrangian is given by L = 12 m l ² x₁² - m g l ( 1 - cos x )

According\; to\; the\; Eulerian - Lagrangian \;equation ,

\frac{d}{dt}\;[\; \frac{\partial }{\partial x_{1}}( L )\;] \; = \; \frac{\partial (L)}{\partial x}

Accordingly , we get \frac{d}{dt} \;[\;\frac{1}{2}\; . 2 \; .\; m\; l^{2}\;x_{1}\;] = m\;g\;l\;sin x ------------ 1

Let x₂ stand for the second derivative of x w.r.t time .

So we derived that m\;l^{2}\;x_{2} = - \;m\;g\;l\;sin x

Sorry I didn't include the minus sign in eqn 1.

or x₂ = - gl x

But x₂ is the accelaration of the system .

So t = 2\pi \;\sqrt{\frac{length}{|accelaration|}} = 2\pi \;\sqrt{\frac{x}{\frac{g}{l}x}} = 2\pi \;\sqrt{\frac{l}{g}}

You don't require any fancy names. What you are asking could be easily obtained by energy conservation: The total mechanical energy of the bob+Earth system remains conserved (why?)
So we get at any \theta,
mg\ell(1-\cos\theta)+\dfrac{1}{2}m\ell^2 \dot{\theta}^2=\mathrm{const}
Differentiating w.r.t. time, we get
g\sin\theta \dot{\theta}+\ell \dot{\theta}\ddot{\theta}=0\quad \Rightarrow\ \ddot{\theta}+\dfrac{g}{\ell}\sin\theta=0
Using small angle approximation and using ω² = g/l, it finally reduces to the shm equation
\ddot{\theta}+\omega^2\theta=0

Generally, a time derivative (i.e. differentiation w.r.t. time) is denoted by putting a dot over that variable.

kkkkkkk

this was a nice question by saumya