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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?

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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.

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  • $\begingroup$ Will you please explain more about this statement: This choice is inconvenient because the derivative of the sin and cosine function would get 2π factors. $\endgroup$
    – arun
    Commented Nov 27, 2011 at 10:03
  • $\begingroup$ @arun: when you use radian measure, the derivative of sin(x) is cos(x), so that near x=0, for small values of $e$, $\sin(e)= e$ , up to corrections which are of size $e^2$. For example, if you take a calculator and find sin(.01) you get .00995.., which is almost .01. If you use turn units, so that the sin function from x=0 to x=1 repeats itself between 1 and 2, and between 2 and 3, instead of from $x=0$ to $x=2\pi$, the sin function is changing a factor of $2\pi$ faster, so that for small x, $\sin(e) = 2\pi e$. The easy derivatives is why radians are preferred units. $\endgroup$
    – Ron Maimon
    Commented Nov 27, 2011 at 22:26
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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.

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    $\begingroup$ It's there a reason for the downvote? $\endgroup$
    – Colin K
    Commented Nov 26, 2011 at 18:36
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    $\begingroup$ Perhaps it was a mistake--- not my downvote. +1. $\endgroup$
    – Ron Maimon
    Commented Nov 28, 2011 at 0:13
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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 $~\sin\theta=b$.

enter image description here

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.

enter image description here

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 $1$.

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