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Does 1 kHz sine tone means $\sin(2(1000)\pi t)$ or $\sin(2(500)\pi t)$?

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    $\begingroup$ Oh man, did we really need 3 answers to answer this??? $\endgroup$ – Matt L. Dec 1 '18 at 11:40
  • $\begingroup$ @MattL. why not four ? ;-) $\endgroup$ – Fat32 Dec 1 '18 at 12:44
  • $\begingroup$ @MattL. But note that one answer does not answer the question "$\sin(2(1000)\pi t)$ or $\sin(2(500)\pi t)$" at all. $\endgroup$ – Dilip Sarwate Dec 1 '18 at 20:00
  • $\begingroup$ @DilipSarwate reading you comment I thought it was me, as I recognized that I used cos rather than the sin function :-)) But I see that it's the other one that does not mention whether it's $500\pi$ or $1000\pi$ ;-) $\endgroup$ – Fat32 Dec 1 '18 at 21:59
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The trigonometric functions "do not know" what a Hertz is and they do not care either. The only thing they know is that a full circle is $2 \pi$ radians. Whether this circle concludes in days, hours, picoseconds or a slice of it represents the angle a force is applied to some lever, is immaterial.

$2 \pi \omega$ expressed in Hertz, denotes a rate. A rate of going around a circle at the time span of a second. $y = \cos(2 \pi 1 t)$ where $t$ is in seconds, would have concluded 1 circle, composed of $2 \pi$ radians, by the time $t$ ticks to 1.

To make it conclude the circle faster, we multiply the "passing of time" (denoted by $t$) by some number $f$.

Therefore, a 1kHz tone is $2 \pi 1000$ radians per second.

Hope this helps.

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  • $\begingroup$ Not sure why the downvote ... I'll undo it. $\endgroup$ – Matt L. Dec 1 '18 at 10:36
  • $\begingroup$ Me neither. I did the same. $\endgroup$ – Cedron Dawg Dec 1 '18 at 14:41
  • $\begingroup$ @MattL. & cedrondawg Thank you for letting me know. I just came back and discover a surprising commotion around this question over the weekend :) $\endgroup$ – A_A Dec 3 '18 at 8:15
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$1$ kHz denotes the frequency, i.e. the inverse of the period of the signal. You have $T=0.001$ seconds and as the period of the sinusoid is $2\pi$,

$$2\pi\cdot1000\cdot T=2\pi.$$

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When the angle $\theta$ of the trigonometric function $\sin(\theta)$ spans a $2\pi$ range, it makes one revolution and to make $f_0$ revolutions in one second (i.e., $f_0$ Hz), the angle should span $2\pi f_0$ range for $t \in [0,1]$, whose mathematical expression will be:

$$ x(t) = \sin( \omega_0 t) = \sin( 2 \pi f_0 t) .$$

With your particular example $f_0 = 1000$ Hz (1k Hz), then you have: $$ x(t) = \sin( \omega_0 t) = \sin( 2 \pi (1000) t) .$$

Note that for simplicity, the relation between the angular frequency $\omega$ in radians (per second) and the frequency $f$ in Hertz is:

$$ \boxed{ \omega = 2 \pi f} $$

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  • $\begingroup$ Your last equation can be expressed completely in units (not dimensions) as: $$ \frac{radians}{second} = \frac{radians}{cycle} \cdot \frac{cycles}{second} $$ $\endgroup$ – Cedron Dawg Dec 1 '18 at 14:42
  • $\begingroup$ @CedronDawg That's very nice. I belive you shoud also add this comment to other answers too. It will be useful for their readers as well. $\endgroup$ – Fat32 Dec 1 '18 at 21:54
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    $\begingroup$ But they didn't state the equation nearly as clearly as you did. I gave you an upvote. $\endgroup$ – Cedron Dawg Dec 1 '18 at 23:19
  • $\begingroup$ @CedronDawg hmm thanks for the upvote ;-) $\endgroup$ – Fat32 Dec 1 '18 at 23:48

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