Sinusoid with period as a linear function of time

Question

I have tried many different methods but to no avail. I am trying to construct a sinusoidal function (e.g. sin(2πf*(x-a))) with a linearly changing period. This is my thought process:

P(t) = a + bt

f(t) = 1/P(t)

We can use the frequency to construct a sinusoid:

y = sin(2π f(t) (t-a))

However, the resulting function does not seem to have a linearly increasing period. Admittedly, my method for checking this could be flawed as I used values from Desmos to check several dP/dt. I have also tried using the integral of 2pi*f inside my sine function, but with the same result. What could I be doing wrong?

Answer 1

So you are constructing functions in the form

$$f(t) = \sin \left(2\pi\cdot\frac{t-a}{a+bt}\right)$$

But when $t\to \infty$, the fraction has a limit $1/b$, so $f$ no longer crosses zero for large enough $t$ but converges to a constant.

When limited in small $t$ though it may work, for example $\sin(2\pi (t-1)/(1+.1t))$ in WolframAlpha:

• for $t\le 10$, which still oscillates,
• for $t\le 1000$, which is more apparent that $f$ no longer crosses zero for large enough $t$.