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

Correct Answer is : f(t)=1ωsinωt

f(t)=\frac{1}{\omega^{2}}(1-\sin \omega t)

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

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

f(t)=\frac{1}{\omega} \sin \omega t

GATE Mechanical Engineering (ME) 2020 Shift 1 Solved Paper

Question : 28 of 65

Marks: +1, -0

The Laplace transform of a function

f(t)

L(f)=\;{1}/{(s^{2}+ω^{2})} .

Then,

f(t)

Solution:

{Lf}( {t})=\;{1}/{ {S}^{2}+ω^{2}}

L^{-1}(\;{1}/{S^{2}+ω^{2}})=f(t)

\;\text " L sin at "\;=\;{a}/{S^{2}+a^{2}}

∴ L \sin ω t=\;{ω}/{S^{2}+ω^{2}}

∴ L^{-1}(\;{1}/{S^{2}+ω^{2}})=\;{\sin ω t}/{ω}

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