Vibrations of Two‑Mass, Two‑Spring System (Wall–k–m1–k–m2)

Derive the frequencies and periods of the two normal modes for two masses m1 and m2 joined by equal springs k with the right spring fixed to a wall. Givens: two springs both of constant k; m1 connected to the wall by the right spring and to m2 by the left spring. Let x1 and x2 be displacements of m1 and m2 to the right. Then the linearized equations of motion are m1 x¨1 = −2k x1 + k x2 m2 x¨2 = k x1 − k x2 Solve the secular equation to get the two eigenfrequencies ω1,2 and hence the periods T1,2. **Answer:** The squared angular frequencies are \[ \omega_{1,2}^2=\frac{k}{2m_1m_2}\Big(2m_2+m_1\pm\sqrt{m_1^2+4m_2^2}\;\Big) \] Therefore the corresponding periods are \[ T_{1,2}=2\pi\sqrt{\frac{2m_1m_2}{k\big(2m_2+m_1\pm\sqrt{m_1^2+4m_2^2}\big)}}\,. \] (For the special case m1=m2=m this reduces to \(\omega^2=\dfrac{k}{2m}(3\pm\sqrt5)\), so \(T=2\pi\sqrt{\dfrac{2m}{k(3\pm\sqrt5)}}\).)