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

Question

TuteeHUB · Accepted Answer

$\;f\left( t \right) = \frac{1}{{{\omega ^2}}}\left( {1 - \sin \omega t} \right)$

TuteeHUB · Answer

$f\left( t \right) = \frac{1}{{{\omega ^2}}}\left( {1 - \cos \omega t} \right)$

TuteeHUB · Answer

$f\left( t \right) = \frac{1}{\omega }\cos \omega t$

TuteeHUB · Answer

$f\left( t \right) = \frac{1}{\omega }\sin \omega t\;$

1.	The Laplace transform of function f(t) is L(t) \( = \frac{1}{{\left( {{s^2} + {\omega ^2}} \right)}}\). Then, f(t) is
A.	\(f\left( t \right) = \frac{1}{{{\omega ^2}}}\left( {1 - \cos \omega t} \right)\)
B.	\(f\left( t \right) = \frac{1}{\omega }\cos \omega t\)
C.	\(f\left( t \right) = \frac{1}{\omega }\sin \omega t\;\)
D.	\(\;f\left( t \right) = \frac{1}{{{\omega ^2}}}\left( {1 - \sin \omega t} \right)\)
Answer» D. \(\;f\left( t \right) = \frac{1}{{{\omega ^2}}}\left( {1 - \sin \omega t} \right)\)

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