= Solution
Let $\delta=x_2-x_1$. The inextensible-string constraint gives the upward displacement $y_2=l-\sqrt{l^2-\delta^2}=\delta^2/(2l)+O(\delta^4/l^3)$. Thus the quadratic <potential energy> is $U=kx_1^2/2+mg(x_2-x_1)^2/(2l)$. The vertical <velocity> of the second <mass> is of second order in the small motion, so the quadratic <kinetic energy> is $m(\dot x_1^2+\dot x_2^2)/2$.
The <Euler-Lagrange equations> for this quadratic <Lagrangian> give
$$
\boxed{\ddot x_1=-\left(\frac km+\frac gl\right)x_1+\frac glx_2,\qquad
\ddot x_2=\frac glx_1-\frac glx_2.}
$$
Small <displacements> are used both in the quadratic approximation to the <gravitational potential energy> and in neglecting the higher-order vertical <kinetic energy>. These are the <spring-supported pendulum normal modes> at linear order. The corresponding force picture sets the string <tension> to $mg$ at leading order and its horizontal angle to $\delta/l$.
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