The Wave Equation — Question 10

Question 10 – Solution

Strategy. A force at the natural frequency produces growing amplitude while the same local wave equation retains its original speed.

Step 1: Verify the resonant amplitude equation. Substitution gives $q''+\omega^2q=(F/\mu)\cos\omega t$. If $D=F/(2\mu\omega)$ and $q=Dt\sin\omega t$, then $$q'=D(\sin\omega t+\omega t\cos\omega t),\qquad q''+\omega^2q=2D\omega\cos\omega t.$$ Thus the proposed response satisfies the forced equation.

Step 2: Check the full initial-boundary data. One has $q(0)=q'(0)=0$. The spatial sine vanishes at $0,L$, giving both fixed ends for all time. The growing factor is a resonant response to sustained external forcing. It changes the amplitude history, not the coefficient $c^2=\mathcal T/\mu$ that sets the source-free propagation speed.

Step 3: Measure energy and work. Spatial integration gives $$\boxed{E(t)=\frac{\mu L}{4}(q'^2+\omega^2q^2) =\frac{F^2L}{16\mu\omega^2} \left[(\sin\omega t+\omega t\cos\omega t)^2 +(\omega t\sin\omega t)^2\right].}$$ Differentiating the first expression and using the amplitude equation yields $E'=(FL/2)q'\cos\omega t$, exactly $\int_0^L F\sin(kx)\cos(\omega t)\,u_t\,dx$. At $t_j=j\pi/\omega$, $$\boxed{E(t_j)=\frac{F^2L}{16\mu}t_j^2.}$$ Energy is unbounded along these times. Its instantaneous rate can have either sign; sustained resonant input produces the growing envelope.

Step 4: Take the frequency limit carefully. The supplied $q_\Omega$ has zero value and derivative at zero. Direct differentiation gives $q_\Omega''+\omega^2q_\Omega=(F/\mu)\cos\Omega t$. For fixed $t$, differentiating numerator and denominator with respect to $\Omega$ at $\omega$ gives $$\boxed{\lim_{\Omega\to\omega}q_\Omega(t) =\frac{F}{\mu}\frac{-t\sin\omega t}{-2\omega} =\frac{Ft\sin\omega t}{2\mu\omega}.}$$ Each fixed nonresonant response is bounded in time, but those bounds are not uniform as its frequency approaches resonance.