Oscillatory motion As I was studying about simple harmonic motion(example pendulum), then I came up to a sin graph as well as a formula that is y = sin2πt/T. I then taken the example of pendulum to understand as to how they have came up to that graph(I understood the graph as to how they would have came across). But I am wondering of how they came up to this formula or I can say to this relationship y = sin2πt/T? Can anyone explain me with diagrams and proofs?
 A: From the definition of simple harmonic motion, that is,

In mechanics and physics, simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium position.(definition as per Wikipedia)

We can say that $F\propto -x$, where if F is force, x is displacement from mean position and if F is torque (though torque is usually denoted by $\tau$, for now, assume F to denote the angular counterpart of force) then x would be angular displacement from the mean position. In both cases, $F \propto \frac{{\rm d^2}{x}}{{\rm d}t^2}$, i.e, force is directly proportional to acceleration or torque is directly proportional to angular acceleration. All of this knowledge would lead to a compact equation i.e.,
$$\frac{{\rm d^2}{x}}{{\rm d}t^2}= -\omega^2x\\$$
(where $\omega^2$ becomes a positive constant, with $\omega$'s value turning out to be equivalent to $\frac{2\pi}{T}$ where T denotes the time period of oscillation). Form the above equation, I'll go on deriving the expression for one directional simple harmonic motion, along direction $x$ with $v$ denoting it's instantaneous velocity. You can do the same for the angular counterpart of the SHM (which is applicable for your pendulum). In the end, the equations would run parallel.
$$v\frac{{\rm d}{v}}{{\rm d}x}=-\omega^2x\\
\int_{v_0}^{v}v {\rm d}v=-\omega^2\int_{x_0}^{x}x{\rm d}x\\ 
\frac{v^2-v_0^2}{2}=-\omega^2\left(\frac{x^2-x_0^2}{2}\right)\\
v^2=v_0^2-\omega^2\left(x^2-x_0^2\right)\\
v^2=\omega^2\left[\left(\frac{v_0^2}{\omega^2}+x_0^2\right)-x^2\right]\\
Let \space A=\sqrt{\left(\frac{v_0^2}{\omega^2}+x_0^2\right)}\\
v^2=\omega^2\left(A^2-x^2\right)\\
v=\omega\sqrt{A^2-x^2}\\
\frac{{\rm d}{x}}{{\rm d}t}=\omega\sqrt{A^2-x^2}\\
\int_{x_0}^{x}\frac{{\rm d}x}{\sqrt{A^2-x^2}}=\omega\int_0^t{\rm d}t \\
\left [\sin^{-1}(\frac{x}{A})\right]_{x_0}^{x}=\omega t\\
Let \space \alpha=\sin^{-1}(\frac{x_0}{A})\\
\sin^{-1}(\frac{x}{A})=\omega t + \alpha\\
x=A\sin(\omega t + \alpha) $$
Since, sine function ranges from -1 to +1, $A$ denotes the amplitude of the displacement. Which means, if the body carries out just the simple harmonic motion, it can move (to the maximum) either A units forward or A units backwards the mean position. With $\omega= \frac{2\pi}{T}$ and $\alpha$ being the initial phase difference, hence is derived the general equation of SHM motion.
For pendulum, the equation converts to
$$\theta=\theta_0\sin(\omega t+ \alpha)$$
Where $\theta$ is the angular displacement and $\theta_0$ is the amplitude of angular displacement. All the other symbols would have the same meaning.
A: This is a standard derivation, which assumes knowledge of calculus. It can easily be found online. I summed up the essence below:
For a pendulum, consider the following sketch:

The displacement of the pendulum bob is $L\theta$, so the acceleration is $L\ddot \theta$ where $\ddot \theta$ is the angular acceleration (second derivative of $\theta$). From there, using Newton's second law, you obtain:
$$\ddot \theta = \dfrac gL \sin\theta$$
The above differential equation is difficult to solve, so we use an approximation -- namely, let $\theta$ be small, such that $\sin\theta\approx\theta$, which yields,
$$\ddot \theta = \dfrac gL \theta$$
The solution to the above differential equation is (I'm not showing the steps -- that is more of a math/basic differential equation solving question than a physics question) $$\theta=\theta_0\cos\left(\omega t +\phi\right)$$ where $\omega=\sqrt{\dfrac {g}{L}}$ and $\phi$ is a phase shift (simply offsets your starting position). Using the fact that $\omega=2\pi f$ and that $T=\dfrac1f$, we get,
$$\theta=\theta_0\cos\left(\dfrac{2\pi}{T}t +\phi\right)$$
By setting the phase $\phi=-\pi/2$ you get a sine curve,
$$\theta=\theta_0\sin\left(\dfrac{2\pi}{T}t +\phi\right)$$
A: To understand why simple harmonic motion is described by $y=\sin \left( \frac {2 \pi t}{T} \right) $ you need to know some calculus.
If you differentiate $y=\sin \left( \frac {2 \pi t}{T} \right) $ twice you get
$$\frac{d^2y}{dt^2} = -\frac{4 \pi^2}{T^2} \sin \left( \frac {2 \pi t}{T} \right) $$
which we can write as
$$\frac{d^2y}{dt^2} = -\omega^2 y $$
where $\omega = \frac{2 \pi}{T}$
Suppose $y$ is the displacement of an object with mass $m$ from an equilibrium position. What force $F$ is required to produce simple harmonic motion about the equilibrium position ? We know that
$$F = ma$$
where $a$ is the acceleration of the object. But $a=\frac{d^2y}{dt^2}$, so we have
$$F = m\frac{d^2y}{dt^2} = -m \omega^2 y$$
In other words, the force $F(y)$ that produces simple harmonic motion is proportional to $y$ in magnitude (because $m$ and $\omega$ are constants) but is in the opposite direction to $y$.
Reversing this chain of reasoning we can see that $y=\sin \left( \frac {2 \pi t}{T} \right)$ is the equation of motion of any object that is subject to a restoring force $F(y) = -ky$ for some positive constant $k$. This force relationship is a common scenario in physics - it occurs in the motion of a weight on a spring (as a result of Hooke's law); a simple pendulum (as long as the maximum angle is small); and the rotational oscillations of an object suspended by a wire.
