A mock`-` up of a `CO_(2)` molecule consists of three balls interconnected by identical light springs and placed along a straight line in the state of equilibrium. Such a system can freely perform oscillation of two types, as shown by the arrows in figure. Knowing the masses of the atomes, find the ration of frequencies of these oscillations

Explanation

In the first mode the carbon atom remains fixed and the oxygent atoms move in equal `&` opposite steps. Then total energy is
`(1)` `(1)/(2) 2 m_(0)dot(x^(2))+(1)/(2) 2 k x ^(2)`
where `x` is the displacement of one of the o atome `(` say left one `).` Thus
`omega_(1)^(2)=k//m_(0)`
`(2)`
In this mode the oxygen atoms move in equal steps in the same direction but the carbon atome moves in such a way as to keep the centre of mass fixed.
Thus `2m_(0)x+m_(c)y=0` or , `y=-(2m_(0))/(m_(c))x`
`K.E.=(1)/(2)2 m_(0)dot(x^(2))+(1)/(2) m_(c)((2m_(0))/(m_(c))dot(x))^(2)=(1)/(2) 2 m_(0)dot(x^(2))+(1)/(2) 2 m_(0)(2m_(0))/(m_(c))dot(x^(2))= (1)/(2) 2 m_(0)(1+(2m_(0))/(m_(c)))dot(x^(2))`
`P.E.=(1)/(2) k(1+(2m_(0))/(m_(c)))x^(2)` `+(1)/(2)k(1+(2m_(0))/(m_(c)))^(2)x^(2)=(1)/(2)2 k (1+(2m_(0))/(m_(c)))^(2)x^(2)`
Thus `omega_(2)^(2)=(k)/(m_(0))(1+(2m_(0))/(m_(c))` and `omega_(2)=omega_(1)sqrt(1+(2m_(0))/(m_(c)))`
Hence, `omega_(2)=omega_(1)sqrt(1+(32)/(12))=omega_(1)sqrt((11)/(3))=1.91omega_(1)`