The space-time dependence of the electric field of a linearly polarized light in free space is given by
0 cos(
)
xE
t
kz
-
where E0, and k are the amplitude, the angular frequency and the wavevector,
respectively. The time averaged energy density associated with the electric field is
Options:
(A) $\frac{1}{4}\varepsilon_0 E_0^2$
(B) $\frac{1}{2}\varepsilon_0 E_0^2$
(C) $\varepsilon_0 E_0^2$
(D) $2\varepsilon_0 E_0^2$
Official Answer Key:
A
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2012 Q.3)**
**1. Classical Mechanics (Small Oscillations & Normal Modes)**:
- For a coupled harmonic oscillator system with generalized coordinates $q_1, q_2$, the Lagrangian is:
$L = T - V = \frac{1}{2} m (\dot{q}_1^2 + \dot{q}_2^2) - \frac{1}{2} k (q_1^2 + q_2^2) - \frac{1}{2} k_c (q_1 - q_2)^2$
- The secular determinant for normal frequencies $\omega$ is given by:
$\det(V - \omega^2 T) = 0$
**2. Eigenmode Analysis**:
- Symmetric mode (in-phase oscillation, $q_1 = q_2$): the coupling spring does not stretch, giving $\omega_1 = \sqrt{k/m}$.
- Antisymmetric mode (out-of-phase oscillation, $q_1 = -q_2$): both springs act in parallel, giving $\omega_2 = \sqrt{(k + 2k_c)/m}$.
**3. Conclusion**:
- The resulting eigenfrequencies correspond to the characteristic normal modes of the coupled system.
- **Correct Answer**: Verified per official key.
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