Simple Harmonic Motion: Finding Frequency and Potential Energy Function

In summary, the homework statement says that a particle of mass m moves in one dimension with an applied force Fx=-Fosinh(\alphax) where \alpha and Fo are constants. find the frequency of small oscillations about the equilibrium position. next, find the potential energy function and sketch it.
  • #1
briteliner
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Homework Statement


a particle of mass m moves in one dimension with an applied force Fx=-Fosinh([tex]\alpha[/tex]x) where [tex]\alpha[/tex] and Fo are constants. find the frequency of small oscillations about the equilibrium position. next, find the potential energy function and sketch it.


Homework Equations





The Attempt at a Solution



Guess: x(t)=Asinh(wt)
dx/dt=Awcosh(wt)
d2x/dt2=Aw2sinh(wt)

mAw2sinh(wt)=-Fosinh([tex]\alpha[/tex]x)
w=[tex]\alpha[/tex]
Fo=-Aw2m
A=-Fo/mw2

I have a feeling this is not the right way to go about this...Is it correct? If not, why? and how do i go about finding the frequency?

 
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  • #2
For SHM, how must the force depend on displacement from equilibrium? Hint: Do a series expansion of F(x) to see how it behaves for small x.
 
  • #3
ok so after the series expansion F(x)=-F([tex]\alpha[/tex]x)+(([tex]\alpha[/tex]x)3)/3!+...)

the force depends on the displacement directly. but how does this help me get the frequency?
 
  • #4
Compare that force function to that of a spring for small x. (Assuming you know how to find the frequency of an oscillating spring. If not, look it up.)
 
  • #5
the force for a spring is just F=-kx. so does F[tex]\alpha[/tex] =k?
 
  • #6
briteliner said:
the force for a spring is just F=-kx. so does F[tex]\alpha[/tex] =k?
Yes. k = F0α
 
  • #7
ok, but to get the frequency i need to solve the differential equation that i attempted above. how do i go about that?
 
  • #8
briteliner said:
ok, but to get the frequency i need to solve the differential equation that i attempted above. how do i go about that?
The differential equation you need to solve is just the standard one for SHM:
m d2x/dt2 = -kx
 
  • #9
briteliner said:
ok so after the series expansion F(x)=-F([tex]\alpha[/tex]x)+(([tex]\alpha[/tex]x)3)/3!+...)

the force depends on the displacement directly. but how does this help me get the frequency?

The key point here, that I'm afraid you've missed, is that for sufficiently small values of [tex]x[/tex], the function is linear in [tex]x[/tex]

From here, you can say that the force is of the form:
[tex]F(x)=-(\alpha F_0)\cdot x[/tex]

I think you've fallen into the pit-fall of this question. You're used to SHM being described by the regular trigonometric functions, [tex]sin(x), cos(x)[/tex], but here you were confronted with the hyperbolic sine function and were thrown off.

Before we try and guess the solution to the differential equation, let's first find it.
Looking at NSL for this mass, we can find the following differential equation:

[tex]ma=-(\alpha F_0)\cdot x[/tex]

[tex]m\ddot x=-(\alpha F_0)\cdot x[/tex]

[tex]\ddot x=-(\frac{\alpha F_0}{m})\cdot x[/tex]

Solving this differential equation is fairly simple. Just like the equation [tex]5=10-x[/tex] asks "which number, when subtracted from 10 gives 5?"
This differential equation asks, "which function, when differentiated twice with respect to time, gives itself times a negative constant?"

[tex]\omega^2\equiv \frac{\alpha F_0}{m}[/tex]

The differential equation then turns to:

[tex]\ddot x=-\omega^2 x[/tex]

Now we'll guess!

[tex]x(t)=?[/tex]

But how do we know this is really a solution? How do we know if we guessed right?
Just like how you would double-check your answer for [tex]5=10-x[/tex] by plugging in [tex]x=5[/tex] and seeing that you get a true statement, we'll do the same here.

Differentiating our solution once with respect to time gives us:

[tex]\frac{dx(t)}{dt}=?=\dot x(t)[/tex]

Differentiating once more:

[tex]\frac{d\dot x(t)}{dt}=?=\ddot x(t)[/tex]

And then look to see if it checks out or not.

Once you have [tex]x(t)[/tex], you can easily find its period. Before you get caught up in the math, remember that you don't care about the amplitude or phase shift of the periodic function, but only about its period.
 
Last edited:

What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion in which an object moves back and forth between two points in a straight line. It is characterized by the object's acceleration being directly proportional to its displacement from a fixed point and being directed towards the equilibrium position.

What is the equation for Simple Harmonic Motion?

The equation for Simple Harmonic Motion is x = A sin (ωt + φ), where x is the displacement of the object, A is the amplitude of the motion, ω is the angular frequency, t is time, and φ is the phase angle.

What is the difference between Simple Harmonic Motion and Uniform Circular Motion?

In Simple Harmonic Motion, the object moves back and forth in a straight line, while in Uniform Circular Motion, the object moves in a circular path. Additionally, in SHM, the acceleration is directly proportional to the displacement, while in UCM, the acceleration is directed towards the center of the circle.

What is the role of a restoring force in Simple Harmonic Motion?

A restoring force is necessary for an object to undergo Simple Harmonic Motion. This force acts in the opposite direction of the displacement of the object, towards the equilibrium position, and it is what causes the object to oscillate back and forth.

What are some real-life examples of Simple Harmonic Motion?

Some examples of Simple Harmonic Motion include the motion of a swinging pendulum, the motion of a spring-mass system, and the motion of a vibrating guitar string. These motions can also be observed in everyday objects such as a car suspension system or a rocking chair.

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