Question

Beats. Consider a free undamped oscillator with mass m=1 and stiffness k=3025, which satisfies the differential equation x′′(t)+3025x(t)=0 The natural frequency is ω0=k/m=55 and the general solution is xc(t)=c1cos(55t)+ c2sin(55t). Now suppose we subject this oscillator to a periodic external force with amplitude 500 and frequency 45 : x′′(t)+3025x(t)=500cos(45t) (a) Find a particular solution of the form xp(t)=Acos(45t)+Bsin(45t). (b) Find the general solution x(t)=xc(t)+xp(t). (c) Find the unique solution x(t) with initial conditions x(0)=0 and x′(0)=0. (d) Express your solution in the form x(t)=Csin(αt)sin(βt). [Hint: Use the trig identities cos(α−β)cos(α+β)cos(α−β)−cos(α+β)=cosαcosβ+sinαsinβ,=cosαcosβ−sinαsinβ,=2sinαsinβ.]

Answer 1

To solve the given forced oscillation problem, we find a particular solution that matches the external force's frequency, then combine it with the general solution and apply initial conditions to find the unique solution, which can then be expressed using a trigonometric identity involving sine functions.

Step-by-step explanation:

When we are given a differential equation representing a forced oscillation as x''(t) + 3025x(t) = 500cos(45t), with mass m=1 and stiffness k=3025, and wish to find a particular solution for this system, we look for a solution of the form xp(t) = Acos(45t) + Bsin(45t). This form assumes that the steady state solution is a simple harmonic motion that matches the frequency of the external force. Plugging this assumed solution into the differential equation, we can solve for A and B. Once we have found A and B, we can combine it with the complementary solution xc(t) = c1cos(55t) + c2sin(55t) to form the general solution.

To find the unique solution satisfying the initial conditions x(0)=0 and x'(0)=0, we need to determine the constants c1 and c2 by applying these conditions to the general solution. Lastly, we can use trigonometric identities to express the solution in the form x(t) = Csin(αt)sin(βt), as hinted in the problem.