Question #88574
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.
1
Expert's answer
2019-05-03T09:16:52-0400

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.


ω1=glω_1= \sqrt{\frac {g} {l}}

ω1=gl+2kmω_1= \sqrt{\frac {g} {l}+\frac {2k} {m}}


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