Sorry! There is no way I can know your background and what is the purpose for your post. There are students just post their homework and want someone to answer it for them.

Since you are a graduate student, I am going to assume your are aware of Tayler's expansion.
We can define a potential for an equilibrium system (SHM is a small deviation from an an equilibrium).
U(x)=U(x0)+ dU/dx|[sub]x=x0[/sub] (x-x0) + (1/2!) d[sup]2[/sup]U/dx[sup]2[/sup]|[sub]x=x0[/sub] (x-x0)[sup]2[/sup]+(1/3!)d[sup]3[/sup]U/dx[sup]3[/sup]|[sub]x=x0[/sub] (x-x0)[sup]3[/sup]+ ...
F=-dU/dx ,so

Fx=- dU/dx|[sub]x=x0[/sub] - d[sup]2[/sup]U/dx[sup]2[/sup]|[sub]x=x0[/sub]*(x-x0) -(1/2!)d[sup]3[/sup]U/dx[sup]3[/sup]|[sub]x=x0[/sub] (x-x0)[sup]2[/sup]-...
  =- d[sup]2[/sup]U/dx[sup]2[/sup]|[sub]x=x0[/sub]*(x-x0) -(1/2!)d[sup]3[/sup]U/dx[sup]3[/sup]|[sub]x=x0[/sub] (x-x0)[sup]2[/sup]-...
(because dU/dx|[sub]x=x0[/sub]=0 at equilibrium point x0)

If the higher order term is smaller compared to the first term,
the above equation reduced to Fx=- d[sup]2[/sup]U/dx[sup]2[/sup]|[sub]x=x0[/sub]*(x-x0) = -k *(x-x0)
That is the reason why a small deviation from the equilibrium will show SHM motion if the higher order term is smaller(can be neglect).

For a small wind, the leave on the tree will show SHM motion.
For a stronger wind, branch of the tree will show SHM motion.
For a hugh wind, the whole tree might show SHM motion.

Since you are a graduate student, I will leave the rest to you to think about it. And you will learn something from it. ;)

Thank you for tell me there is something with the spell check function.