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Problem:

Let $f(t) = \sin(\omega_1 t)+\sin(\omega_2 t)$.

What condition must $\omega_1$ and $\omega_2$ satisfy for $x(t)$ to be periodic? When $x(t)$ is periodic, what is its period?

My Attempt:

Let $f(t)$ be periodic.

Let the time period of $f(t)$, $\sin(\omega_1 t)$ and $\sin(\omega_2 t)$ be $T$, $T_1$ and $T_2$ respectively.

Since $\sin$ has period $2\pi$, $\sin(\omega t) = \sin(\omega_1 t + 2\pi) = \sin(\omega_1(t + \frac{2\pi}{\omega_1}))$.

So $T_1 = \frac{2\pi}{\omega_1}$ and by the same logic $T_2 = \frac{2\pi}{\omega_2}$.

So

$\begin{aligned}\\f(t) &= f(t+T) \\ &= \sin(\omega_1 (t + T)) + \sin(\omega_2 (t + T)) \\& = \sin(\omega_1 t + \omega_1 T) + \sin(\omega_2 t + \omega_2 T) \\ &= \sin(\omega_1 t + 2\pi) + \sin(\omega_2 t + 2\pi).\\\end{aligned}$

Not really sure that this gets me anywhere...

As far as I am aware in order for $f(t)$ to be periodic then the time periods of $\sin(\omega_1 t)$ and $\sin(\omega_2 t)$ must have a rational LCM, so $f(x)$ is periodic if there exist $a$ and $b$ such that $a T_1 = b T_2 = r$. Is this correct? If so how can I show this?

Thanks in advance for your help!

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    $\begingroup$ Hint: If $f$ is periodic with period $T$ and has a second derivative, then $f''$ is also periodic with period $T$. Use this to prove that if $\omega_1^2 \ne \omega_2^2$, then the period of $f$ is also a period for both sines. $\endgroup$ Aug 6, 2017 at 20:23
  • $\begingroup$ @Gribouillis That is really clever. Why don't you make this an answer, so that I can upvote it ;) $\endgroup$ Aug 6, 2017 at 20:37
  • $\begingroup$ @SeverinSchraven Done! $\endgroup$ Aug 6, 2017 at 20:45

2 Answers 2

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Hint

If $f$ is periodic with period $T$ and has a second derivative, then $f''$ is also periodic with period $T$. Use this to prove that if $\omega_1^2 \ne \omega_2^2$, then the period of $f$ is also a period for both sines

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Note that $\omega_1 T$ and $\omega_2 T$ are not necessarily equal to $2\pi$. In fact, $\omega_1 T = 2\pi k_1$, and $\omega_2 T = 2\pi k_2$ for some positive integers $k_1$ and $k_2$. Since $\omega_1 = \frac{2\pi}{T_1}$, and $\omega_2 = \frac{2\pi}{T_2}$, $\omega_1 T = 2\pi k_1$, and $\omega_2 T = 2\pi k_2$ imply $T = k_1 T_1$, and $T = k_2 T_2$. Lastly, we have to find the smallest $T$ that is a multiple of both $T_1$ and $T_2$. Does LCM ring any bell?

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