Mechanical Vibrations — Question 6

Question 6 – Solution

Strategy. The homogeneous correction needed to fit rest data creates the nearby-frequency interference; exact resonance requires a different particular form.

Step 1: Solve the nonresonant problem. Newton’s law gives $x''+25x=2\cos 4t$, $x(0)=x'(0)=0$. Since $25-16=9$, a particular response is $(2/9)\cos 4t$. Fitting zero data gives $$\boxed{x=\tfrac 29(\cos 4t-\cos 5t).}$$ Its second derivative plus $25x$ equals $2\cos 4t$, and both initial data vanish.

Step 2: Separate the two time scales. The cosine-difference identity gives $$\boxed{x=\tfrac 49\sin(9t/2)\sin(t/2).}$$ The carrier angular frequency is $9/2\,\mathrm{s^{-1}}$. The nonnegative envelope is $A(t)=(4/9)|\sin(t/2)|\,\mathrm m$, with period $2\pi\,\mathrm s$. Its zeros occur at $t=2n\pi$, and its largest value is $4/9\,\mathrm m$, attained by $x$ at $t=\pi$. The envelope period uses the absolute value; the signed slow sine has period $4\pi$.

Step 3: Distinguish a rest state from remaining at rest. Here $x'=\tfrac 29(-4\sin 4t+5\sin 5t)$, so $x(2\pi)=x'(2\pi)=0$. Yet the applied force then equals $2\,\mathrm N$; the equation gives $x''(2\pi)=2\,\mathrm{m/s^2}$. The mass immediately accelerates again. Reaching the equilibrium rest state while a nonzero force persists does not keep it there.

Step 4: Solve at exact resonance. For $x''+25x=2\cos 5t$, use $x_p=At\sin 5t$. Its residual is $10A\cos 5t$, so $A=1/5$ and the data are already zero: $$\boxed{x_{\mathrm{res}}=\tfrac 15t\sin 5t.}$$ At $t_n=(\pi/2+2n\pi)/5$, it equals $t_n/5\to\infty$, whereas the beating response is bounded by $4/9$. Beats arise from interference of distinct frequencies; resonance produces a growing factor when the forcing matches the natural frequency.