The Laplace transform of function f(t) is L(t)$=\frac{1}{\left(s^2+{\omega }^2\right)}$. Then, f(t) is

The correct answer is:
$f(t)=\frac{1}{\omega }\sin \omega t$

Step-by-Step Explanation:

We are given the Laplace transform of a function $f(t)$, denoted as $L(s)$:
$L(s)=\frac{1}{s^2+{\omega }^2}$
We need to find the inverse Laplace transform, $f(t)={\mathcal{L}}^{-1}\{L(s)\}$.

Recall the standard Laplace transform formula for the sine function:
$\mathcal{L}\{\sin (\omega t)\}=\frac{\omega }{s^2+{\omega }^2}$

Taking the inverse Laplace transform of both sides of this standard relation gives:
${\mathcal{L}}^{-1}(\frac{\omega }{s^2}$