Finding the Period of a Function Confusion

Sep 2012
66
0
Scotland
Hi,

I have this function, y(t) = 8sin(5PIt-2)

I'm asked to find the period which is 0.4 and I know this is worked out by 2/5.

But I'm wondering why is it done this way? Would you not usually do this - 2PI/5PI - 2?

Help is much appreciated, thanks.
 
Jun 2008
1,389
513
Illinois
The general form of a sinusoidal function is \(\displaystyle y(t) = A \sin (\omega t + \phi)\), where $\omega $ is the angular velocity of the function in radians/second and $\phi$ is the phase angle, which is essentially the offset of the function. You can think of \(\displaystyle \phi\) as shifting the sine wave to the left or the right, but it has no effect on the "speed" with which the sine wave moves up and down.

The period for the sinusoidal function is \(\displaystyle \frac {2 \pi} {\omega}\), and the frequency in hertz (cycles per second) is \(\displaystyle f = \frac 1 T = \frac {\omega} {2 \pi}\)$.
 
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Sep 2012
66
0
Scotland
Ah right ok, I get it now. I misunderstood the period equation, thanks.