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

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

• Oh man, did we really need 3 answers to answer this??? – Matt L. Dec 1 '18 at 11:40
• @MattL. why not four ? ;-) – Fat32 Dec 1 '18 at 12:44
• @MattL. But note that one answer does not answer the question "$\sin(2(1000)\pi t)$ or $\sin(2(500)\pi t)$" at all. – Dilip Sarwate Dec 1 '18 at 20:00
• @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 $1000\pi$ ;-) – Fat32 Dec 1 '18 at 21:59

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.

• Not sure why the downvote ... I'll undo it. – Matt L. Dec 1 '18 at 10:36
• Me neither. I did the same. – Cedron Dawg Dec 1 '18 at 14:41
• @MattL. & cedrondawg Thank you for letting me know. I just came back and discover a surprising commotion around this question over the weekend :) – A_A Dec 3 '18 at 8:15

$$1$$ 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 $$2\pi$$,

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

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:

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

• Your last equation can be expressed completely in units (not dimensions) as: $$\frac{radians}{second} = \frac{radians}{cycle} \cdot \frac{cycles}{second}$$ – Cedron Dawg Dec 1 '18 at 14:42
• @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. – Fat32 Dec 1 '18 at 21:54
• But they didn't state the equation nearly as clearly as you did. I gave you an upvote. – Cedron Dawg Dec 1 '18 at 23:19