Mechanical Vibrations — Question 7

Question 7 – Solution

Strategy. Different measured quantities introduce different frequency factors; optimize their amplitudes separately.

Step 1: Match harmonic coefficients. Set $x=A\cos\omega t+B\sin\omega t$. The equation $2x''+4x'+20x=F_0\cos\omega t$ gives $$(20-2\omega^2)A+4\omega B=F_0,\qquad -4\omega A+(20-2\omega^2)B=0.$$ Writing $\Delta=(20-2\omega^2)^2+16\omega^2>0$, we obtain $$A=\frac{F_0(20-2\omega^2)}{\Delta},\quad B=\frac{4F_0\omega}{\Delta},\quad \boxed{X=\frac{F_0}{\sqrt\Delta},\quad V=\omega X.}$$ The homogeneous roots are $-1\pm 3i$, so all initial-data transients decay and this periodic response is the long-time motion.

Step 2: Maximize displacement. Complete the square: $$\Delta=4[(\omega^2-8)^2+36].$$ It has its unique minimum for $\omega>0$ at $\omega_X=2\sqrt 2\,\mathrm{s^{-1}}$. Therefore $\boxed{X_{\max}=F_0/12}$ in meters when $F_0$ is expressed in newtons.

Step 3: Maximize velocity. Divide the amplitude denominator by $\omega$: $$V=\frac{F_0}{\sqrt{(20/\omega-2\omega)^2+16}}.$$ This is largest exactly when $20/\omega-2\omega=0$, giving $$\boxed{\omega_V=\sqrt{10}\,\mathrm{s^{-1}},\qquad V_{\max}=F_0/4}$$ in meters per second when $F_0$ is in newtons. The spring and inertial contributions to this denominator cancel there.

Step 4: Interpret the phase and the two maxima. At $\omega=\sqrt{10}$, $A=0$ and $B=F_0/(4\sqrt{10})>0$. Thus $x=[F_0/(4\sqrt{10})]\sin(\sqrt{10}t)$, a displacement lag of $\pi/2$ behind the cosine force. Its displacement amplitude is smaller than $F_0/12$, even though its velocity amplitude is largest. The displacement peak, velocity peak and undamped natural frequency need not all coincide; the response quantity must be named.