Question 4.7.6: The upper end of a flexible cable is supported by a stationa......

With the FBD indicated in Figure 4E6b, considering the vertical forces for the elementary length dy of the cable, one obtains

(T +dT)  cos  ( θ  +  \frac{\partial θ}{\partial y}   dy  )   =  Tcosθ+μgdy,

where μ is the length density in mass per unit length, and θ is the angular displacement as indicated in Figure 4E6b.
For small oscillations, the angular displacement θ is small such that the above equation reduces to

dT =μgdy     or     T =μgy .                                (i)

Now, consider forces along the horizontal direction by using Newton’s law of motion (that is, along x here):

(μdy)  \frac{\partial²  u}{\partial t²}   =  (T +dT)  sin  (θ  +\frac{\partial θ}{\partial y}   dy  ) −Tsinθ.

For small oscillations,

(μdy)  \frac{\partial²  u}{\partial t²}   =    (T +dT)   (θ  +\frac{\partial θ}{\partial y}   dy )  −Tθ

(μdy)  \frac{\partial²  u}{\partial t²}   =  Tθ  +  T  \frac{\partial θ}{\partial y}   dy   +  d Tθ  +  dT  \frac{\partial θ}{\partial y}  dy  −Tθ

(μdy)  \frac{\partial²  u}{\partial t²}   =  T  \frac{\partial θ}{\partial y}   dy   +  (θ  +\frac{\partial θ}{\partial y}   dy )  dT  . 

Substituting for Equation (i), the above equation becomes

(μdy)  \frac{\partial²  u}{\partial t²}   =  T  \frac{\partial θ}{\partial y}   dy   +  (θ  +\frac{\partial θ}{\partial y}   dy )  μgdy  . 

It may be appropriate to note that at this stage one should not delete the common factor dy on both sides of the equation. If such action is taken one would arrive at an erroneous equation. In order to obtain the correct equation of motion, one should expand the rhs of the last equation and re-arrange terms to give

(μdy)  \frac{\partial²  u}{\partial t²}   =  T  \frac{\partial θ}{\partial y}   dy   +  (θμg)  dy   +    \frac{\partial θ}{\partial y}   (dy)²  μg  .

Since only small oscillations are considered, the higher-order term associated with (dy)² can be disregarded such that the above equation reduces to

(μdy)  \frac{\partial²  u}{\partial t²}   =    T  \frac{\partial θ}{\partial y}   dy    +  (θμg)  dy

Now, the common factor dy may be canceled on both sides so that it becomes

μ  \frac{\partial²  u}{\partial t²}   =  T  \frac{\partial θ}{\partial y}   +  (θμg) .                      (ii)

From Equation (i), T = μgy and recall that θ = \frac{\partial u}{\partial y}  ;  one therefore has

This is the required equation of motion. Note that the rhs of Equation(iii) contains the second order-partial derivative with a time-dependent coefficient.