Mechanical Vibrations — Question 2

Question 2 – Solution

Strategy. Peak spacing determines the oscillation frequency; peak ratios determine exponential decay. A known mass converts these rates into physical coefficients.

Step 1: Infer the coefficients. An underdamped response has form $e^{-\alpha t}(A\cos\omega_dt+B\sin\omega_dt)$, with $\alpha=c/(2m)$ and $\omega_d^2=k/m-\alpha^2$. Same-sign peaks are one damped period apart, so $$\omega_d=\frac{2\pi}{\pi}=2\,\mathrm{s^{-1}},\qquad e^{-\alpha\pi}=e^{-\pi/2}\ \Longrightarrow\ \alpha=\tfrac 12\,\mathrm{s^{-1}}.$$ Thus $\boxed{c=3/2\,\mathrm{N\,s/m}}$ and $\boxed{k=51/8\,\mathrm{N/m}}$. Adjacent positive and negative extrema would be only half a period apart.

Step 2: Reconstruct and verify the peaks. The initial value gives $A=0.08$, and $x'(0)=0$ gives $B=0.02$. Hence $$\boxed{x=0.08e^{-t/2}[\cos 2t+\tfrac 14\sin 2t],} \qquad x'=-0.17e^{-t/2}\sin 2t.$$ The characteristic roots are $-1/2\pm 2i$, verifying the equation. Extrema occur at $t=n\pi/2$; positive maxima occur at $t=n\pi$, where $x=0.08e^{-n\pi/2}$. The derivative changes from positive to negative at each interior positive maximum, and the release at zero starts a decrease.

Step 3: Compare energies. At an extremum, velocity is zero, so $E=kx^2/2$. Successive positive-peak energies therefore have ratio $$\boxed{e^{-\pi}},$$ not $e^{-\pi/2}$. Energy depends quadratically on displacement at a turning point.

Step 4: State the identification limit. Without a known mass, only $c/m=2\alpha=1\,\mathrm{s^{-1}}$ and $k/m=\alpha^2+\omega_d^2=17/4\,\mathrm{s^{-2}}$ are determined. Multiplying $m,c,k$ by the same positive constant leaves the normalized equation, the complete free response, and all peak measurements unchanged. The three coefficients cannot then be identified separately.