Find the frequency of the oscillation of a horizontal beam.

Click For Summary

Homework Help Overview

The problem involves a uniform meter stick pivoted at one end and held horizontal by a spring, with the goal of determining the frequency of its oscillation when displaced slightly. The context includes concepts from dynamics and oscillatory motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to write a torque equation considering the forces acting on the beam and expresses confusion regarding the application of differential equations in this context. They question the influence of gravity on the oscillatory motion and seek clarification on their approach.

Discussion Status

Participants are actively engaging with the problem, with one suggesting that the original poster's equation can be simplified by recognizing that certain terms are constants. There is an acknowledgment of the need to differentiate between the oscillatory motion and the equilibrium position, but no consensus has been reached on the specific solution method.

Contextual Notes

The original poster indicates a limitation in their understanding of differential equations, which may affect their ability to progress in solving the problem. Additionally, the presence of gravity in the system is noted as a complicating factor in their analysis.

Ishaan S
Messages
19
Reaction score
0

Homework Statement



A uniform meter stick of mass M is pivoted on a hinge at one end and held horizontal by a spring with spring constant k attached at the other end. If the stick oscillates up and down slightly, what is its frequency? [Hint: Write a torque equation about the hinge.] The length of the beam is 1.25 m.

Homework Equations



τ = Iα

Fspring = -kx

The Attempt at a Solution



I have found the question online and have linked it to this comment in case someone needs to see a picture of the scenario.

I started by writing a torque equation, with the beam angled θ above the horizontal and x being the displacement of the spring tip above the horizontal.

The torques about the pivot on the wall on the stick, using the small angle approximation, are:

1) The spring exerts a force, -kx = -kLθ , a distance, L, away from the pivot. Thus, the torque exerted by the spring is, for small angles, -kxL2θ.

2) The center of mass exerts a torque, given by -mgL/2.

The torque equation is. therefore, ∑τ = Iθ'' = -kL2θ - mgL/2.

This is where I was stuck. I don't know how to solve differential equations beyond separable ones as of yet. The book I am using shows that the solution to a simple harmonic oscillator's equation of motion is just a sinusoid; however, this does not seem like a simple oscillator as there is gravity in this system, which is odd as this is the simple oscillatory section of my book.

I tried substituting a function of cosine; however it does not seem to work. If anyone could shed light as to what I did incorrectly or what I am overlooking, it would be greatly appreciated.

Thank you in advance.
 

Attachments

Physics news on Phys.org
I forgot to mention, the problem in the attachment is number 5.
 
Consider the function y=sin(x)+47

When you differentiate, the constant will just disappear, so the function will be a solution to y''= -(y-47)

You could do this more generally to find the solution of y'' = -C1y+C2 (which is the form of your equation) but I just wanted to make a point.

mgL/2 is just a constant, so the solution to your equation will be a sinusoid plus a constant. Adding a constant doesn't affect the frequency of oscillation (it just shifts the equilibrium position which it is oscillating about).
 
Ok. Thank you very much. I get it now.
 

Similar threads

Replies
19
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 17 ·
Replies
17
Views
4K
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
12
Views
11K
  • · Replies 15 ·
Replies
15
Views
6K
Replies
6
Views
2K
Replies
1
Views
3K