Sine wave, $\pi$ and frequency Please explain the relation $\sin(2\pi ft)$ such that how the $\pi$ (which is actually circumference/diameter of a circle) relates with the sine wave which is having a longitudinal vibration?
 A: It's just using radian measure for angles. If you have a sin and cosine function, you feed it an angle, and if you choose the angle to be measured in fractions of a turn, then the relation would be $\sin(ft)$ and $\cos(ft)$. This choice is inconvenient because the derivative of the sin and cosine function would get $2\pi$ factors. So instead, people parametrize the angles by radians, so that the derivatives are simple, but the radian measure gives the $2\pi$ in the frequency deinition, so that $sin(\omega t)$ repeats every $2\pi/\omega$ time-units, and the frequency is $f={\omega\over 2\pi}$. The factor $\omega$ is the radian frequency, and it is often just called the frequency in sloppy physics-talk.
A: The sine function has a period of $2 \pi$ radians, and that relates directly to the circle - the value of the function is swept out across one rotation around the circumference of the circle, which represents an angle of $2 \pi$ radians.
Using the knowledge that sine has period $2\pi$, we know that $sin(2\pi f)$ has period f.
So at time t, we'll have a value of the function $sin(2\pi f t)$ - and just set t=1 to get back to f cycles per second.
A: Consider the unit circle.  Let $\theta$ be an arbitrary angle with initial side on the positive x-axis and the vertex at the origin.  Let $(a,b)$ be the coordinates of the point, where the terminal side of the angle intersects the unit circle.
We define $~\sin\theta=b$. 

Now imagine, $\theta$ as a variable.  That is, the terminal side of $\theta$ is free to rotate.  Then you can see that $\sin\theta$ increases from $0$ to $1$ as $\theta$ increases from $0$ to $\frac{\pi}{2}$, decrease from $1$ to $0$ as $\theta$ increases from $\frac{\pi}{2}$ to $\pi$,  decrease from $0$ to $-1$ as $\theta$ increases from $\pi$ to $\frac{3\pi}{2}$, and  increases from $-1$ to $0$ as $\theta$ increases from $\frac{3\pi}{2}$ to $2\pi$.  You may see now that $f(\theta)=\sin\theta$ is a periodic function with period $2\pi$.
If you sketch the graph of $f(\theta)=\sin\theta$ in $\theta f(\theta)$-plane from $\theta=0$ to $\theta=2\pi$, you get the following graph.

If you let $\theta=2\pi t$, then as $\theta$ varies from $0$ to $2\pi$, $t$ varies from $0$ to $1$.  Therefore, the period of the function $f(t)=\sin2\pi t$ is $1$.
