Solution of I.V.P for harmonic oscillator with driving force is given by Inverse Laplace transformy"+ω2y=sin⁡γt,y(0)=0,y′(0)=01) y(t)=L−1(γ(s2+ω2)2)2) y(t)=L−1(γs2+ω2)3) y(t)=L−1(γ(s2+γ2)2)4) y(t)=L−1(γ(s2+γ2)(s2+ω2))

Step 1Given the differential equation for harmonic oscillator with driving force is y"+ω2y=sin⁡γt,y(0)=0,y′(0)=0Step 2Taking Laplace transform on both sidesL{y"+ω2y}=L{sin⁡γt}L{y"}+L{ω2y}=γs2+γ2 L is linear⇒s2L{y}−sy(0)−y′(0)+ω2L{y}=γs2+γ2⇒(s2+ω2)L{y}=γs2+γ2 y(0)=0,y′(0)=0⇒L{y}=γ(s2+γ2)(s2+ω2)Taking inverse Laplace transformy(t)=L−1{γ(s2+γ2)(s2+ω2)}Thus, option 4) is correct

"Solution of I.V.P for harmonic oscillator with driving..." solved and explained

Daniaal Sanchez

Answered question

2021-02-16

Solution of I.V.P for harmonic oscillator with driving force is given by Inverse Laplace transform
$y " + ω^{2} y = \sin γ t, y (0) = 0, y^{'} (0) = 0$
1) $y (t) = L^{- 1} (\frac{γ}{(s^{2} + ω^{2})^{2}})$
2) $y (t) = L^{- 1} (\frac{γ}{s^{2} + ω^{2}})$
3) $y (t) = L^{- 1} (\frac{γ}{(s^{2} + γ^{2})^{2}})$
4) $y (t) = L^{- 1} (\frac{γ}{(s^{2} + γ^{2}) (s^{2} + ω^{2})})$

Answer & Explanation

insonsipthinye

Skilled2021-02-17Added 83 answers

Step 1
Given the differential equation for harmonic oscillator with driving force is

y " + ω^{2} y = \sin γ t, y (0) = 0, y^{'} (0) = 0

Step 2
Taking Laplace transform on both sides

L {y " + ω^{2} y} = L {\sin γ t}

L {y "} + L {ω^{2} y} = \frac{γ}{s^{2} + γ^{2}} L is linear

\Rightarrow s^{2} L {y} - s y (0) - y^{'} (0) + ω^{2} L {y} = \frac{γ}{s^{2} + γ^{2}}

\Rightarrow (s^{2} + ω^{2}) L {y} = \frac{γ}{s^{2} + γ^{2}} y (0) = 0, y^{'} (0) = 0

\Rightarrow L {y} = \frac{γ}{(s^{2} + γ^{2}) (s^{2} + ω^{2})}

Taking inverse Laplace transform

y (t) = L^{- 1} {\frac{γ}{(s^{2} + γ^{2}) (s^{2} + ω^{2})}}

Thus, option 4) is correct

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