= Solution
The squared-frequency matrix is
$$
\frac gl I+\frac km\begin{pmatrix}1&-1&0\\-1&2&-1\\0&-1&1\end{pmatrix}.
$$
The dimensionless spring matrix has an <orthonormal eigenbasis>
$$
v_1=\frac1{\sqrt3}(1,1,1),\quad
v_2=\frac1{\sqrt2}(1,0,-1),\quad
v_3=\frac1{\sqrt6}(1,-2,1),
$$
with <eigenvalues> $0,1,3$. Adding the gravitational term therefore gives
$$
\boxed{\omega_1=\sqrt{\frac gl},\quad
\omega_2=\sqrt{\frac gl+\frac km},\quad
\omega_3=\sqrt{\frac gl+\frac{3k}m}.}
$$
The uniform mode leaves both springs unstretched. The other two modes deform the springs and so have higher frequencies.
Back to article page
