A $1 \mathrm{~kg}$ block on a spring is modeled by the following initial value problem.
\[
x^{\prime \prime}+4 x=12 \cos (4 t), x(0)=0, x^{\prime}(0)=0
\]
a) Find the position of the block at time $t$
\[
x(t)=
\]
b) Assuming the forcing function and the spring stay the same, what mass would the block have to have for the system to be in resonance?
kg. (exact answer)
Answer
Since the spring constant k is 4 (from the \(x'' + 4x = 0\) part of the differential equation), the mass m must be: \(m = 4 / 16 = 0.25 \, kg\). \(\boxed{m = 0.25 \, kg}\)
Step 1 ：The given differential equation is a non-homogeneous second order differential equation with a cosine forcing function. We can solve it using the method of undetermined coefficients.
Step 2 ：The homogeneous solution of the equation is given by the characteristic equation: \(r^2 + 4 = 0\), which gives \(r = \pm2i\). So, the homogeneous solution is: \(x_h(t) = C1 \cos(2t) + C2 \sin(2t)\).
Step 3 ：For the particular solution, we guess a solution of the form: \(x_p(t) = A \cos(4t) + B \sin(4t)\).
Step 4 ：Substituting this into the differential equation gives: \(-16A \cos(4t) - 16B \sin(4t) + 4A \cos(4t) + 4B \sin(4t) = 12 \cos(4t)\).
Step 5 ：Simplifying gives: \(-12A \cos(4t) - 12B \sin(4t) = 12 \cos(4t)\). Comparing coefficients gives A = -1 and B = 0. So, the particular solution is: \(x_p(t) = - \cos(4t)\).
Step 6 ：The general solution is the sum of the homogeneous and particular solutions: \(x(t) = C1 \cos(2t) + C2 \sin(2t) - \cos(4t)\).
Step 7 ：Using the initial conditions \(x(0) = 0\) and \(x'(0) = 0\), we find \(C1 = 0\) and \(C2 = 0\). So, the solution is: \(x(t) = - \cos(4t)\).
Step 8 ：For the system to be in resonance, the natural frequency of the system must equal the frequency of the forcing function. The natural frequency is given by \(\sqrt{k/m}\), where k is the spring constant and m is the mass. The frequency of the forcing function is 4 (from the \(\cos(4t)\) term).
Step 9 ：Setting these equal gives: \(\sqrt{k/m} = 4\). Solving for m gives: \(m = k / 16\).
Step 10 ：Since the spring constant k is 4 (from the \(x'' + 4x = 0\) part of the differential equation), the mass m must be: \(m = 4 / 16 = 0.25 \, kg\). \(\boxed{m = 0.25 \, kg}\)
