Find the Frequency (I think?) of a Sinusoidal Equation

Question

Given equation: $v = .96 \sin (\pi t/3.4)$, where $v$ is the velocity (in liters/sec) of airflow during a respiratory cycle and $t$ is the time in seconds.

I need to find the number of cycles per minutes, in units of cycles per min. Can anyone help? Thanks!

Answer 1

Every time the input parameter of the sine function increases by $2\pi,$ the function goes through another cycle. So from $0$ to $2\pi$ is one cycle, from $0$ to $4\pi$ is two cycles, etc.

In $t$ seconds, the input parameter of the sine in your equation goes from $0$ to $\pi t/3.4.$ Set $t = 60$ and you have the change in the parameter over $60$ seconds. So that gives you a number, and that number is $2\pi$ times the number of cycles.

Answer 2

Sinusoids follow the form $$a\sin(bx+c)$$

• $a$ is the amplitude, since $\sin$ naturally has amplitude $1$
• $2\pi/b$ is the period, since $\sin$ naturally has period $2\pi$ and multiplying $x$ by a constant greater than $1$ causes a horizontal compression
• $c$ tells you the phase shift

You can take it from there.

Answer 3

Hint: Recall that in $a\sin[b(x-h)]+k$, $b$ represents the number of cycles in the domain $0 \leq x \leq 2\pi$ and $\frac{2\pi}{b}$ represents the period, or length of a single cycle.