Pendulum with horizontal spring

[mountainbiker]

Homework Statement

A 5kg sphere is connected to a thin massless but rigid rod of length L=1.3 m to form a simple pendulum. The rod is connected to a nearby vertical wall by a spring with a spring constant k= 75N/m, connected to it at a distance h=1.1 m below the pivot point of the pendulum. What is the angular frequency (in rad/s) of the system for small amplitude oscillations.

Homework Equations

$$\alpha$$ = $$\omega$$²*$$\theta$$
I=m*r²
Torque=f*d*sin$$\theta$$

or small oscillations, sin$$\theta$$=$$\theta$$ and cos$$\theta$$=1

The Attempt at a Solution

sum the torque and set them equal to $$\alpha$$I

gravitational torque= Lmg$$\theta$$

if theta is the angle between the rod and its equilibrium position, the angle of the spring torque is (theta + $$\pi$$/2), so the spring torque is
f*d*sin(theta + $$\pi$$/2). The force is k*$$\Delta$$x, or h*sin(theta + $$\pi$$/2) and the distance is h, giving

torque from spring =k*h²*sin$$\theta$$*sin(theta + $$\pi$$/2)
or k*h²*$$\theta$$*(theta + $$\pi$$/2)

My final equation is

mL²$$\omega$$²tex]\theta[/tex] = Lmg$$\theta$$ + k*h²*$$\theta$$*(theta + $$\pi$$/2)

The problem is that this gives me a $$\omega$$ that is dependent on $$\theta$$, which isn't possible. It should be constant. Can someone please tell me where I'm going wrong?

 

[ehild]

sin(theta+pi/2)= cos(theta). For small angles, cos(theta) =1.

ehild