Answer to Question #88574 in Classical Mechanics for Shivam Nishad

Question

The equations of motion of two coupled
spring-mass systems (of equal masses)
executing longitudinal oscillations are
(d²x1/dt²)+w0²x1–Wa²(x2 — x1 )= 0
and
(d²x/dt²)+w0²x2–ws²(x2–x1)=0
where w0 = √k'/m and ws= √k/m
are the natural frequency and coupling frequency of the oscillators.
Decouple these equations and obtain
expressions for normal mode frequencies.

Answer 1

A normal mode of an oscillating system is the motion in which all parts of the system move sinusoidally with the same frequency and with a fixed phase relation.

$$ω_1= \sqrt{\frac {g} {l}}$$

$$ω_1= \sqrt{\frac {g} {l}+\frac {2k} {m}}$$