Simple harmonic motion of a bar pivoted at one end

In summary: This new equilibrium position of the spring would have a torque from the spring, and an upward torque from the torque on the bar.
  • #1
Martin89
25
1
Homework Statement
See below...
Relevant Equations
Equations of torque and simple harmonic motion
Hi, I am unsure how to proceed with this problem. I believe that I can correctly calculate the frequency of the oscillations for a bar that is not suspended from a spring but I do not know how to take the effect of the spring into account. The answer given by my professor is $$
f=\frac{1}{2\pi}\sqrt{\frac{3\alpha^2k}{M}}$$
Problem.png
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  • #2
Hello Martin,

That's funny: your derivation has no ##k## and profs' has no ##g##

What do you mean when you say torque is ##\alpha l Mg\sin\theta## ?
 
  • #3
##Mgsin\theta## is the force on the bar and the length of the bar is ##\alpha L##, although for a uniform bar the force should act through the centre of mass I believe?
 
  • #4
Martin89 said:
##Mgsin\theta## is the force on the bar and the length of the bar is ##\alpha L##, although for a uniform bar the force should act through the centre of mass I believe?

The force on the bar when displaced must depend on ##k##.
 
  • #5
Martin89 said:
##Mgsin\theta## is the force on the bar and the length of the bar is ##\alpha L##, although for a uniform bar the force should act through the centre of mass I believe?
Yes. So at equilibrium the spring is extended to compensate ##{1\over 2} Mg##.

But what is the restoring torque if the spring is extended a little further ? Express the extension in ##\theta## too and bingo !
 
  • #6
PeroK said:
The force on the bar when displaced must depend on ##k##.

This is the part I'm having difficulty with. I know that the force on the bar must depend on ##k## but I don'nt know how to express it.

Working backwards from my professor's answer I believe that the torque on the rod is given by ##Torque=\alpha^2KL^2\sin\theta##. However, I don't understand why there is no dependence on ##Mg##?
 
  • #7
Martin89 said:
This is the part I'm having difficulty with. I know that the force on the bar must depend on ##k## but I don'nt know how to express it.

Working backwards from my professor's answer I believe that the torque on the rod is given by ##Torque=\alpha^2KL^2\sin\theta##. However, I don't understand why there is no dependence on ##Mg##?

First, write down the equation for the initial equilibrium of the bar. You have a torque from ##Mg## but you must also have an upward torque from the spring. Note that the spring itself, therefore, must initially be stretched.

Now, what happens when you pull the bar down and extend the spring a little further?

Hint: you could take the equilibrium position of the spring itself, hanging under its own weight as ##0##; the equilibrium position of the spring holding the bar as ##x_0## and take the displacement in addition to this, i,e, ##x_0 + x##.

I'm using ##x## here as the downward displacement of the spring.
 

Related to Simple harmonic motion of a bar pivoted at one end

1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which a system or object moves back and forth around a central equilibrium point. The motion is characterized by a constant frequency and amplitude, and can be described using mathematical functions such as sine or cosine.

2. How is simple harmonic motion different from other types of motion?

Simple harmonic motion is different from other types of motion because it is a special case of harmonic motion, where the restoring force is directly proportional to the displacement from equilibrium. This results in a sinusoidal motion, whereas other types of motion may have different shapes or patterns.

3. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is x(t) = A sin(ωt + φ), where x is the displacement from equilibrium, A is the amplitude, ω is the angular frequency, and φ is the phase angle.

4. How does a bar pivoted at one end exhibit simple harmonic motion?

A bar pivoted at one end is an example of a simple pendulum, which exhibits simple harmonic motion because the restoring force (due to gravity) is directly proportional to the displacement of the bar from its equilibrium position. As the bar swings back and forth, the motion follows a sinusoidal pattern.

5. What factors affect the period of simple harmonic motion in a bar pivoted at one end?

The period of simple harmonic motion in a bar pivoted at one end is affected by the length of the bar, the acceleration due to gravity, and the mass of the bar. The longer the bar, the longer the period, while a larger acceleration due to gravity or a heavier bar will result in a shorter period.

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