Kernel: SageMath 9.7

C22 Laplace Transforms Part 3

Rules Summary and Preview.

FunctionLaplace Transformf(t)0estf(t)dteat1/(sa)tnn!/sn+1af(t)+bg(t)aF(s)+bG(s)f(t)sF(s)f(0)cos(bt)s/(s2+b2)sin(bt)b/(s2+b2)eatf(t)F(sa)H(tc)f(tc)ecsF(s)tf(t)F(s) \qquad \def\arraystretch{1.4} \begin{array}{ll} \text{Function} & \text{Laplace Transform}\\ \hline f(t) & \int_{0}^{\infty} e^{-st}f(t) \, dt \\ e^{at} & 1/(s-a) \\ t^n & n!/s^{n+1} \\ af(t) + bg(t) & aF(s) + bG(s) \\ f'(t) & sF(s) - f(0) \\ \cos(bt) & s/(s^2 + b^2) \\ \sin(bt) & b/(s^2 + b^2) \\ e^{at}f(t) & F(s-a) \\ H(t-c)f(t-c) & e^{-cs}F(s) \\ tf(t) & -F'(s) \\ \end{array}

Exercise 5.3.9. An undamped spring-mass-damper system with mass m=2m = 2 kg and spring constant k=8k = 8 newtons per meter is at equilibrium position u=0u = 0 and is not moving at time t=0t = 0. No additional forces act on the mass until time t=10t = 10 seconds, but for t>10t > 10 a force f(t)=40f(t) = 40 newtons is applied to the mass. At time t=15t = 15 seconds the force drops to zero. Find the position of the mass for t>0t > 0 by formulating an appropriate ODE with initial conditions and solving with the Laplace transform. Plot the solution on the interval 0t250 ≤ t ≤ 25.