Mechanical Vibrations — Question 1

Question 1 – Solution

Strategy. Distinguish extension from natural length from displacement about equilibrium, and keep the upward initial velocity negative.

Step 1: Establish the coordinate and model. Write $z=\ell+x$ for downward extension, where $\ell=0.196\,\mathrm m$. The static balance is $k\ell=mg$, so $k=19.6/0.196=100\,\mathrm{N/m}$. Newton’s law gives $$2x''=mg-k(\ell+x)=-100x, \qquad x(0)=0.05,\quad x'(0)=-0.20.$$ Thus $\omega_0=\sqrt{k/m}=5\sqrt 2\,\mathrm{s^{-1}}$.

Step 2: Solve and find the amplitude. Fitting both data gives $$\boxed{x(t)=\frac 1{20}\cos(5\sqrt 2t)-\frac 1{25\sqrt 2}\sin(5\sqrt 2t).}$$ The amplitude satisfies $R^2=x(0)^2+[x'(0)/\omega_0]^2=1/400+1/1250=33/10000$. Hence $\boxed{R=\sqrt{33}/100\,\mathrm m}$. Differentiation verifies $x''+50x=0$ and the two data.

Step 3: Locate the first crossing. The first zero lies in $0<\omega_0t<\pi/2$, where the motion is upward. Solving $x=0$ gives $$\boxed{t_*=\frac 1{5\sqrt 2}\arctan\frac{5\sqrt 2}{4}.}$$ At that crossing, $x'=-\omega_0R=-\sqrt{66}/20\,\mathrm{m/s}$. The negative sign identifies upward travel; using an arbitrary branch of arctangent could miss the first crossing.

Step 4: Check energy and physical extension. Relative to equilibrium, the conserved energy is $E=mx'^2/2+kx^2/2$. Initially, $$E=\tfrac 12(2)(0.20)^2+\tfrac 12(100)(0.05)^2=0.165\,\mathrm J.$$ At $x=0$, $E=x'^2$, giving $|x'|=\sqrt{0.165}=\sqrt{66}/20\,\mathrm{m/s}$. The maximum natural-length extension is $\ell+R=0.196+\sqrt{33}/100\,\mathrm m$, about $0.25345\,\mathrm m$. Gravity has not been neglected: its constant force cancels the spring’s equilibrium force after shifting the coordinate.