Modeling — Question 4

Question 4 – Solution

Strategy. The coupling spring exerts opposite forces on the masses; symmetric and antisymmetric motions then become independent oscillators.

Step 1: Derive the complete state model. The middle spring’s extension is $q_2-q_1$, giving force $q_2-q_1$ on mass 1 and its negative on mass 2. Thus $q_1''=-2q_1+q_2$ and $q_2''=q_1-2q_2$, with numerical acceleration coefficients in s$^{-2}$. For $Z=(q_1,v_1,q_2,v_2)^T$, $$\boxed{Z'=(v_1,-2q_1+q_2,v_2,q_1-2q_2)^T,\quad Z(0)=(a,0,0,0)^T.}$$

Step 2: Solve the two normal modes. Set $s=q_1+q_2$, $d=q_1-q_2$. Then $s''=-s$, $d''=-3d$, with $s(0)=d(0)=a$ and zero initial derivatives. Therefore $$\boxed{q_1=\tfrac a2(\cos t+\cos\sqrt 3t),\quad q_2=\tfrac a2(\cos t-\cos\sqrt 3t).}$$ The velocities are $v_1=-a(\sin t+\sqrt 3\sin\sqrt 3t)/2$ and $v_2=-a(\sin t-\sqrt 3\sin\sqrt 3t)/2$. They satisfy all four initial data.

Step 3: Check the energy balance. In joules, with the stated numerical masses and stiffnesses, $$E=\tfrac 12(v_1^2+v_2^2)+\tfrac 12\bigl(q_1^2+q_2^2+(q_2-q_1)^2\bigr).$$ Its derivative is $v_1(q_1''+2q_1-q_2)+v_2(q_2''+2q_2-q_1)=0$. At the initial state $\boxed{E=a^2\text{ J}}$ when $a$ is the numerical initial displacement in metres. Energy includes the middle spring as well as the wall springs.

Step 4: Test an exact return. A return of the full state requires $s=d=a$ and $s'=d'=0$. Thus $t=2\pi m$ and $\sqrt 3t=2\pi n$ for integers $m,n$. For $t>0$ this would imply $\sqrt 3=n/m$, impossible. No exact positive-time return occurs. A near return visible in a plot cannot prove periodicity.