Step 1
Given that, \(R = 70\).
The given function is, \(F(s) =\frac{s}{R^2s^2+16\pi^2}\)
\(F(s)=\frac{s}{70^2s^2+16\pi^2}\)
\(F(s)=\frac{s}{4900s^2+16\pi^2}\)
\(f(t)=L^{-1}\left(\frac{s}{4900s^2+16\pi^2}\right)\)
It is known that, \(L^{-1}(sF(s))=\frac{d}{(dt)}f(t)+f(0)\)
Step 2
Apply the above property and obtain the inverse Laplace transform of the given function as shown below.
\(f(t)=L^{-1}\left(\frac{s}{4900s^2+16\pi^2}\right)\)
\(f(t)=L^{-1}\left(\frac{1}{4900s^2+16\pi^2}\right)+L^{-1}\left(\frac{1}{4900s^2+16\pi^2}\right)(0)\)
\(f(t)=\frac{d}{(dt)}\left(\frac{1}{280} \sin \left(\frac{(2\pi t)}{35}\right)\right)+0\)
\(f(t)=\frac{1}{4900} \cdot \cos\left(\frac{2\pi t}{35}\right)\)
Thus, the required inverse Laplace transform is obtained.
Compute the Inverse Laplace Transform of F(s) =frac{s}{R^2s^2+16pi^2}take R=70
Answers (1)
Relevant Questions
\(L\left\{t-e^{-3t}\right\}\)
which of the laplace transform is
\(1.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s-3}\)
\(2.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s-3}\)
\(3.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}+\frac{1}{s+3}\)
\(4.)\ L\left\{t-e^{-3t}\right\}=\frac{1}{s^{2}}-\frac{1}{s+3}\)
Solution of I.V.P for harmonic oscillator with driving force is given by Inverse Laplace transform
\(y"+\omega^{2}y=\sin \gamma t , y(0)=0,y'(0)=0\)
1) \(y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\omega^{2})^{2}}\bigg)\)
2) \(y(t)=L^{-1}\bigg(\frac{\gamma}{s^{2}+\omega^{2}}\bigg)\)
3) \(y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})^{2}}\bigg)\)
4) \( y(t)=L^{-1}\bigg(\frac{\gamma}{(s^{2}+\gamma^{2})(s^{2}+\omega^{2})}\bigg)\)
\(g(t)=\begin{cases}0 & 0<t<11 & 1<t<36&3<t<54&5<t\end{cases}\) &0
\(g(t)=\prod_{1,3}(t)+6\prod_{3,5}(t)+4u(t-5)\)
compute the Laplace transform of g(t)
Let x(t) be the solution of the initial-value problem
(a) Find the Laplace transform F(s) of the forcing f(t).
(b) Find the Laplace transform X(s) of the solution x(t).
\(x"+8x'+20x=f(t)\)
\(x(0)=-3\)
\(x'(0)=5\)
\(\text{where the forcing } f(t) \text{ is given by }\)
\(f(t) = \begin{cases} t^2 & \quad \text{for } 0\leq t<2 ,\\ 4e^{2-t} & \quad \text{for } 2\leq t < \infty . \end{cases}\)
