= Solution
To specify “initial motion” unambiguously, choose each mode to pass through equilibrium at $t=0$ with $\dot{\boldsymbol x}(0)$ proportional to its listed <mode shape>. The arrows in the figure then describe relative initial <velocities>:
* In mode 1, all three masses move in the same direction; the central mass has $\sqrt2$ times either outer <velocity>.
* In mode 2, the two outer masses move with equal and opposite <velocities>, while the central mass remains stationary.
* In mode 3, the outer masses move together while the central mass moves oppositely with $\sqrt2$ times their speed.
Reversing every arrow changes only the common <phase>. A quarter-cycle later these same ratios describe the extremal <displacements>, with zero instantaneous <velocity>.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2019-fixed-chain-mode-motions.png]
{title=Relative initial velocities of the three fixed-end normal modes, starting as the masses pass through equilibrium.}
{description=Arrow lengths follow the mode shapes. The antisymmetric mode has equal and opposite outer velocities and zero central velocity; its angular frequency is unchanged by any positive change in central mass.}
Now replace the central mass by $M$, leaving the outer <masses> equal to $m$. The antisymmetric <mode shape> $q=(1,0,-1)^T$ gives zero net force on the central mass, since $k x_1+k x_3=0$. Each outer mass still obeys $m\ddot x=-2kx$. Consequently
$$
\boxed{\omega_2=\sqrt{\frac{2k}{m}}\text{ is unchanged for every }M>0.}
$$
This <middle-mass-independent mode of a fixed three-mass spring chain> is exact, not merely a large-$M$ approximation. The other two modes move the central mass and generally change. For example, their symmetric-sector frequencies satisfy
$$
(2k-m\omega^2)(2k-M\omega^2)-2k^2=0,
$$
which explicitly depends on $M$.
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