This is an interesting question for me:
Find the Laplace transform of $\mathcal{L} \{sin(\sqrt 3t)\}$.
I know the Laplace transform of $sin(\sqrt t) = \frac{\sqrt\pi}{2s^{3/2}}e^{-1/4s}$, but simply multiplying t by 3 really messes things up.
I've looked through the books solution, and they write:
$\mathcal{L} (sin(\sqrt 3t)) = \mathcal{L}(sin2(\sqrt\frac{3}{4}t) = \sqrt \frac{3\pi}{4} e^{-3/4s}{s^{3/2}}$. They say they do this by utilising the Laplace property of $\mathcal{L}\{\frac{sin2\sqrt {at}}{\sqrt {\pi a}}\} = \frac{e^{-a/s}}{s^{3/2}}$, which I don't understand (for example, there's no $\sqrt \pi a$ in the denominator, so what the heck?).
Can anyone please help explain how this works?
Best Answer
I've figured it out.
It's just manipulating the expression, and then multiplying it by $\sqrt \frac{3\pi}{4} $ to negate the denominator in the Laplace transform. Very smart.