1
$\begingroup$

I know that $f = \frac{\text{cycles}}{\text{unit of time}}$ and I know that period $T = \text{time it takes to complete one cycle}$.

So if we take the definition

$f = \frac{1}{T} = \frac{\text{1 cycle}}{\text{time it takes to complete one cycle}}$

So if we have some $x$ cycles we get that $f = \frac{x \text{ cycles}}{\text{time it takes to complete one cycle}}$.

How can we have $x > 1$ if it takes $T$ to complete a cycle? In other words, how could $x > 1$ have completed in the time it takes to complete one cycle?

I’m guessing that $T$ is somehow decoupled from $x$ and I’ve got my definitions mixed up.

$\endgroup$
  • $\begingroup$ In your second equation for $f$, the $1$ has no units. So, $f$ has units $[\text{cycle}/\text{s}]$ (cycles per second) and $T$ has units $[\text{s}/\text{cycle}]$ (seconds per cycle). Also, note that it is very common not to use cycles as a unit and define Hz as $1/s$ (or $s^{-1}$). $\endgroup$ – MBaz Aug 30 '18 at 22:26
2
$\begingroup$

It's about units and dimensions. And I'm probably not the guy to cast an answer on that but I would like to state it as follows. In short; length, time, mass etc. are dimensions, whereas meter, seconds, kg are units of them respectively in SI. (for other dimensions and units, a physics book might help.)

Now, if a wheel rotates 7 times in 2 seconds; it's rotation frequency $f$ is found as: $$ f = \frac{7 ~\text{ cycles}}{2 ~\text{ seconds}} = 3.5 ~\text{cycles-per-second (cps)}.$$

As can be seen, frequency is given by a (cycle) number divided by time. The cycle number has no physical units or dimensions; it's just a repetition count. Whereas the duration time has a unit of seconds (in SI).

Thus the dimesion of frequency is 1/time, whose SI unit is $1/s = s^{-1}$. This unit is also, and more frequently, denoted as Hz where it (implicitly) stands for (cycles) per second indeed. So, $3.5$ Hz stands for $3.5$ cycles per second.

Now, similarly the period is defined as the duration time per one cycle. Again note that since cycle is a number and does not posses a unit, then the dimension of the period is time and its unit is seconds in SI. But you can always denote it also as seconds per cycle, rarely practiced though.

Finally, using both notations you have $$ T = \frac{1}{f} $$

So if frequency $f$ is defined with a unit of Hz ($s^{-1}$), then the period will have a unit of seconds (s) in SI. Otherwise if you define frequency as cycles per second then the period unit will be seconds per cycle.

So if you define the frequency as $3.5$ cycles-per-second , then the period associated is: $$T = \frac{1}{f} = \frac{1}{ 3.5 \frac{\text{cyc}}{s} } = \frac{2}{7} \frac{\text{s}}{\text{cyc}}$$

Otherwise, in accepted norm, as was you would define freqquency as $3.5$ Hz. its period in SI will be: $$ \boxed{ T = \frac{1}{f} = \frac{1}{ 3.5 s^{-1} } = \frac{2}{7} ~\text{ s} } $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.