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

In summary, the rod oscillates for a period of time determined by the equation of motion dependent on x, but not on \phi.
  • #1
LauraPhysics
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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 [tex]\phi[/tex]

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

(d^2[tex]\phi[/tex])/dt^2=-(3k/m)sin[tex]\phi[/tex]cos[tex]\phi[/tex]-(3g/2L)sin[tex]\phi[/tex] (10 marks

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

Homework Equations



d^2x/dt^2 +[tex]\omega[/tex]^2=0

F=-kx (spring)

F=ma

F= -mgsin[tex]\phi[/tex] (pendulum)

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

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 [tex]\phi[/tex].

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

Then used the small angle approx to make sin[tex]\phi[/tex] equivalent to tan [tex]\phi[/tex]

Tan[tex]\phi[/tex] 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.
 
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  • #2
LauraPhysics said:
So far, I have been able to deduce an equation of motion dependent on x, but not on [tex]\phi[/tex].

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

I calculated the restoring force for both the pendulum and the spring, I then used Newton's 2nd law to get -kx-mgsin[tex]\phi[/tex]=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?
 
  • #3
What about part B?!
 

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

What is a physical pendulum attached to a spring in SHM?

A physical pendulum attached to a spring in SHM is a system that consists of a rigid body pendulum (such as a rod or a bar) attached to a spring, which is then fixed at one end. When the pendulum is displaced from its equilibrium position, it experiences a restoring force from the spring, causing it to oscillate back and forth in a periodic motion.

What is the equation of motion for this system?

The equation of motion for a physical pendulum attached to a spring in SHM is given by:

mL2θ" + kL(sinθ - θcosθ) = 0

Where m is the mass of the pendulum, L is the length of the pendulum, k is the spring constant, and θ is the angular displacement from equilibrium.

How is the period of oscillation affected by the parameters of the system?

The period of oscillation of a physical pendulum attached to a spring in SHM is affected by the mass of the pendulum, the length of the pendulum, and the spring constant. A larger mass or longer length will result in a longer period, while a larger spring constant will result in a shorter period.

What is the relationship between the amplitude and the energy of the system?

The amplitude of the motion of a physical pendulum attached to a spring in SHM is directly proportional to the energy of the system. As the amplitude increases, so does the energy of the system. This is because the potential energy stored in the spring increases with the amplitude, and this energy is converted to kinetic energy as the pendulum oscillates.

Can the equation of motion be applied to any initial conditions?

Yes, the equation of motion for a physical pendulum attached to a spring in SHM can be applied to any initial conditions. This is because the equation takes into account the initial angular displacement, as well as the mass, length, and spring constant of the system. It can also be solved for different scenarios, such as when the pendulum is released from rest or when it is given an initial velocity.

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