Physical Pendulum attached to spring in SHM. Find Equation of Motion

[LauraPhysics]

Homework Statement

Uniform Rod of length 0.2m and mass 0.2kg pivoted at one end. THe other end of attached to a horizontal spring with spring constant 3.0N/m. The spring is neither stretched nor compressed when the rod is perfectly vertical. You can also assume that the force due to the spring is always horizontal. The angle between the rod and equilibrium shall be called \phi

A) Show that the equation of motion for the rod is :

(d^2\phi)/dt^2=-(3k/m)sin\phicos\phi-(3g/2L)sin\phi (10 marks

B) Determine the rod's oscillation period in the small-angle approximation

Homework Equations

d^2x/dt^2 +\omega^2=0

F=-kx (spring)

F=ma

F= -mgsin\phi (pendulum)

d^2\phi/dt^2= (-Mgl/I)\phi

I= 1/3ML^2

The Attempt at a Solution

So far, I have been able to deduce an equation of motion dependent on x, but not on \phi.

I calculated the restoring force for both the pendulum and the spring, I then used Newton's 2nd law to get -kx-mgsin\phi=md^2x/dt^2

Then used the small angle approx to make sin\phi equivalent to tan \phi

Tan\phi is x/l which subs into Eqn as -kx-mgx/l=md^2x/dt^2

Divide through by m and this leaves -k/mx-gx/l = d^2x/dt^2

Would be very grateful for any help with the where to go from here.

 

[naresh]

So far, I have been able to deduce an equation of motion dependent on x, but not on \phi.

You might want to use the fact that if \phi is in radians and is small, sin \phi can be approximated to \phi.

I calculated the restoring force for both the pendulum and the spring, I then used Newton's 2nd law to get -kx-mgsin\phi=md^2x/dt^2

This is not quite correct. The problem asks you to assume that the spring restoring force is horizontal. Along what direction are you using Newton's second law? Moreover, there are forces at the pivot that you have not considered in this equation.

Think about what happens physically. Which part of the rod moves back and forth? Is it linear motion?

 

[Cyberslayer]

What about part B?!