The figure shows a system of two equal masses m and three massless horizontal
springs with spring constants k1 , k2 , k1. Ignore gravity. The masses can move only
in the horizontal direction and there is no dissipation. If m = 1, k1 = 2 and k2 = 3
(all in appropriate units), the frequencies of the normal modes of the system in the
same system of units are
Options:
(A) √2 , √8
(B) √2 , √6
(C) √3 , √10
(D) √3 , √8
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2025 Q.38)**
**1. Coupled Oscillations & Normal Modes**:
- Two identical masses $m$ connected by three springs with spring constants $k_1, k_2, k_1$.
- Displacements of the two masses: $x_1(t)$ and $x_2(t)$.
**2. Equations of Motion & Normal Mode Eigenvalues**:
- Kinetic and potential energy of the coupled system:
$T = \frac{1}{2}m\dot{x}_1^2 + \frac{1}{2}m\dot{x}_2^2$
$V = \frac{1}{2}k_1 x_1^2 + \frac{1}{2}k_2(x_2 - x_1)^2 + \frac{1}{2}k_1 x_2^2$
- Equations of motion:
$m\ddot{x}_1 + (k_1 + k_2)x_1 - k_2 x_2 = 0$
$m\ddot{x}_2 + (k_1 + k_2)x_2 - k_2 x_1 = 0$
- Setting $x_1 = A_1 e^{i\omega t}, x_2 = A_2 e^{i\omega t}$:
$\det\begin{pmatrix} -m\omega^2 + k_1 + k_2 & -k_2 \\ -k_2 & -m\omega^2 + k_1 + k_2 \end{pmatrix} = 0$
- Normal mode frequencies:
- Symmetric mode ($x_1 = x_2$):
$\omega_1 = \sqrt{\frac{k_1}{m}}$
- Antisymmetric mode ($x_1 = -x_2$):
$\omega_2 = \sqrt{\frac{k_1 + 2k_2}{m}}$
- Substitute given parameters $k_1/m = 2$ and $k_2/m = 3$:
$\omega_1 = \sqrt{2}$
$\omega_2 = \sqrt{2 + 2(3)} = \sqrt{2 + 6} = \sqrt{8}$
- The normal mode angular frequencies are **$\sqrt{2}, \sqrt{8}$**.
**3. Official Verification**:
The verified official IIT answer is **Option (A)** (`√2 , √8`).
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