2. At time t the displacement from equilibrium, y(t), of an...

2. At time $t$ the displacement from equilibrium, $y(t)$, of an undamped spring-mass system of mass $m$ is governed by the initial-value problem egin{align*} frac{d^2y}{dt^2} + omega^2y = frac{F_0}{m} cos omega t, quad y(0) = 1, quad frac{dy}{dt}(0) = 0 end{align*} where $F_0$ and $omega$ are positive constants. Solve this initial value problem to determine the motion of the system. What happens as $t o infty$

Transcript

00:01 In this problem we are provided with the differential equation d squared y over d t squared plus omega squared y equals to f0 over m cause of omega t we are provided with the initial conditions y of 0 equals to 1 and d y over d t of 0 equals to 0.

00:26 So first let us consider the characteristic equation.

00:29 So here we have m squared plus omega squared to be equal to 0 which implies that m squared equals to negative omega squared taking square root on both sides m equals to plus or minus omega times i.

00:45 So the corresponding general solution will be y equals to c1 times cause of omega t plus c2 times sine of omega t so now by the method of undetermined coefficients let us consider the particular solution to be a times t times cost of omega t plus b times t times sine of omega t which can be rewritten as t times a cost omega t plus b sine omega t now we substitute this particular solution in the given differential equation and find out the answer for a and b.

01:33 So in order to substitute it, we require the second derivative.

01:37 So let us differentiate the particular solution.

01:41 We have, we make use of the product rule of differentiation.

01:46 So we get a times cost of omega t plus b times sine of omega t plus t times negative a omega, sine of omega t, plus b times omega, cost of omega t.

02:05 Now let us differentiate this again.

02:08 We have yp double dash to be equal to we have negative a omega sine of omega t plus b omega cos of omega t and again applying the product rule here.

02:25 We have negative a omega sine of omega t plus b omega cosa of omega t and now letting t remain as it is and differentiating the two terms we have negative a omega squared cos omega t minus b times omega squared sine omega t so now let us substitute this in the given differential equation.

03:00 We get negative 2 times a times omega times sine omega t plus 2b times omega times cos omega t minus a omega square t cos omega t minus b omega square t times sine of omega t.

03:24 T plus a times t times omega squared times cos of omega t plus b t omega squared sine of omega t equals to f not over m cause of omega t so here we see that these two terms cancel out these two terms cancel out as well and we are left with negative 2a omega sine of omega t plus 2b omega cost of omega t equals to f0 over m cause of omega t.

04:05 By equating the coefficient of sine of omega t on both sides, we see that the value of a equals to 0.

04:13 Likewise, equating the coefficient of cost of omega t on both sides, we get 2 times b times omega equals to f0 over m, which implies that b equals to f0 over 2 times m times omega.

04:33 So this is the required value of a and b.

04:35 Let us substitute it in the equation of yp...

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