Discussion
A mass spring system consists of two blocks of mass M and one block of mass m (\(\lt\)M). The blocks are connected with two springs of same spring constant (k), as shown in figure. The system is constraint to move along a straight line on a frictionless horizontal surface. Let V represent the potential energy of the system. Considering that the masses are exhibiting small oscillations, the potential energy matrix (V) is given as :
Answer: A
The potential energy matrix is \(\frac12\begin{pmatrix}k & -k & 0\\ -k & 2k & -k\\ 0 & -k & k\end{pmatrix}\).
With displacements \(x_{1}, x_{2}, x_{3}\) of the three blocks, each spring stores energy according to its own stretch:
\[ \begin{aligned} V &= \tfrac{1}{2}k(x_{1} - x_{2})^{2} + \tfrac{1}{2}k(x_{2} - x_{3})^{2} \\ &= \tfrac{1}{2}(kx_{1}^{2} + 2kx_{2}^{2} + kx_{3}^{2}) \\ &\quad - kx_{1}x_{2} - kx_{2}x_{3} \end{aligned} \]
Writing V as \(\frac{1}{2}x^{T}Kx\) puts k, 2k, k on the diagonal and -k for each pair joined by a spring, with 0 for the end blocks, which are not directly connected. The matrix must be symmetric with negative off-diagonal couplings, which rules out the other three. The masses do not enter V; they belong to the kinetic energy matrix.
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