Finding the Right Answer: Tips for Solving Oscillation Problems

In summary, the conversation is about learning oscillation and understanding the relationship between force and acceleration in a simple harmonic oscillator. The participants discuss the equation for acceleration and the complementary solution to the differential equation, as well as the importance of the negative sign in the force equation. They also mention the use of angular frequency and its advantages in advanced studies.
  • #1
xtrubambinoxpr
87
0
I apologize ahead of time for all of these post about oscillation. I am trying to learn this stuff on my own.

ImageUploadedByPhysics Forums1389919648.734460.jpg


I answered "a" because the x^2 function. But I don't know the second part.. Would it also be 3x^2 because it is proportional to the x(t) function?
 
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  • #2
What type of force acts in a SHO? What is the dependence on displacement?
 
  • #3
nasu said:
What type of force acts in a SHO? What is the dependence on displacement?

depends on the force being acted on it?
 
  • #4
What depends on the force? I don't understand this.

Imagine a simple harmonic oscillator. You displace it from equilibrium by a distance x.
What force acts to bring it back to equilibrium? What is the dependence of this force of x?
You can think about a specific system, like a ball attached to a spring. In this case the force is the elastic force isn't it?
 
  • #5
The gravity would be pulling it down and the force would be opposite.
Or with a block the spring is applying the force if it's on a table both ways. Acceleration will be applied opposite of the motion. So the force is given by the spring
 
  • #6
xtrubambinoxpr said:
Or with a block the spring is applying the force if it's on a table both ways. Acceleration will be applied opposite of the motion. So the force is given by the spring
Good. If the spring stretches by x, what is the force that the spring exerts on the mass? From Newton's second law, what is the acceleration (as a function of the spring constant, the mass of the mass, and the displacement from equilibrium x)?
 
  • #7
Chestermiller said:
Good. If the spring stretches by x, what is the force that the spring exerts on the mass? From Newton's second law, what is the acceleration (as a function of the spring constant, the mass of the mass, and the displacement from equilibrium x)?
You lost me. Only using the spring formulas or the oscillation ones? I don't see the connection. If F=kx. Then kx/m=a
 
  • #8
xtrubambinoxpr said:
You lost me. Only using the spring formulas or the oscillation ones? I don't see the connection. If F=kx. Then kx/m=a
Actually, it looks like you do see that connection. You just didn't realize it. These are the equations I was asking about. Actually, the equation for the acceleration should have a minus sign, because the spring is a restoring force constantly trying to bring the mass back to the equilibrium position; but the mass always overshoots the equilibrium position. Back to the equation for the acceleration: a = -kx/m. Look at your problem statement now, and see which relationship between the acceleration and the displacement agrees with this equation.

Now, back to the relationship between this and SHM. The acceleration is the second derivative of the displacement with respect to time, so we have:
[tex]a=\frac{d^2x}{dt^2}=-\frac{k}{m}x[/tex]
The complimentary solution to this differential equation is:

x=A sin(ωt) + Bcos(ωt)

where [itex]ω=\sqrt{\frac{k}{m}}[/itex]

The constants A and B in the equation depend on the initial displacement and the initial velocity of the mass.

Do you see now how this ties in with SHM?

Chet
 
  • #9
Chestermiller said:
Actually, it looks like you do see that connection. You just didn't realize it. These are the equations I was asking about. Actually, the equation for the acceleration should have a minus sign, because the spring is a restoring force constantly trying to bring the mass back to the equilibrium position; but the mass always overshoots the equilibrium position. Back to the equation for the acceleration: a = -kx/m. Look at your problem statement now, and see which relationship between the acceleration and the displacement agrees with this equation.

Now, back to the relationship between this and SHM. The acceleration is the second derivative of the displacement with respect to time, so we have:
[tex]a=\frac{d^2x}{dt^2}=-\frac{k}{m}x[/tex]
The complimentary solution to this differential equation is:

x=A sin(ωt) + Bcos(ωt)

where [itex]ω=\sqrt{\frac{k}{m}}[/itex]

The constants A and B in the equation depend on the initial displacement and the initial velocity of the mass.

Do you see now how this ties in with SHM?

Chet

So, if I understand this completely, I was wrong because there is no x^2 in regards to the acceleration. it would be the -4x option because it seems to be closely related to what we just proved. I think I was thinking of ω[itex]^{2}[/itex]
 
  • #10
xtrubambinoxpr said:
You lost me. Only using the spring formulas or the oscillation ones? I don't see the connection. If F=kx. Then kx/m=a

There's a very important minus sign missing in your Force formula.

Have you been set this question without any prior information about SHM and do you have no text to work from?
 
  • #11
sophiecentaur said:
There's a very important minus sign missing in your Force formula.

Have you been set this question without any prior information about SHM and do you have no text to work from?

I have the text and I now remember the (-) sign in the equation. Completely my mistake. I have 2 books I am trying to learn from so forgive me if I forget something.
 
  • #12
Your book (or Wiki) will have it all in there - you may just need to read it more slowly.

What is the relationship (always) between force and acceleration and how can you relate that to your Force equation (with the right sign included). Now look at those alternatives in the question and find one that has the same form as this.

PS Using angular frequency ω, is always annoying and perplexing to start with but it is very useful to get used to it because, once you start to get advanced, if you continue to use frequency f, the constant 2∏ keeps cropping up, multiple times and just gets in the way.
 
  • #13
sophiecentaur said:
Your book (or Wiki) will have it all in there - you may just need to read it more slowly.

What is the relationship (always) between force and acceleration and how can you relate that to your Force equation (with the right sign included). Now look at those alternatives in the question and find one that has the same form as this.

PS Using angular frequency ω, is always annoying and perplexing to start with but it is very useful to get used to it because, once you start to get advanced, if you continue to use frequency f, the constant 2∏ keeps cropping up, multiple times and just gets in the way.

based on what you're saying now I think it is either 5x or -4x. the force is directly proportional to the acceleration. I think I am missing the final key point
 
  • #14
By the way if anyone were to want to skype or google hangout on this topic I would be cool with that too. Just trying to learn.
 
  • #15
Why was I so picky about the sign? It tells you the direction that the force is acting in and is essential in the mathematical description of the motion. Which direction is the force? To reduce or increase the magnitude of displacement?
 
  • #16
sophiecentaur said:
Why was I so picky about the sign? It tells you the direction that the force is acting in and is essential in the mathematical description of the motion. Which direction is the force? To reduce or increase the magnitude of displacement?

it is negative because it is the restoring force to get it back to its initial position
 
  • #17
So which of the alternatives is showing that?
 
  • #18
sophiecentaur said:
So which of the alternatives is showing that?

-4x is the only one abiding by that principle
 
  • #19
So, how does that look? A reasonable answer?
 
  • #20
sophiecentaur said:
So, how does that look? A reasonable answer?

being that it needs to have the equation form to fit, yes I say it is reasonable. a(t) = -ω[itex]^{2}[/itex]x(t).. that being said -4x is the only option that seems to fit.
 
  • #21
xtrubambinoxpr said:
being that it needs to have the equation form to fit, yes I say it is reasonable. a(t) = -ω[itex]^{2}[/itex]x(t).. that being said -4x is the only option that seems to fit.

That is a fresh addition. Can you explain where it came from? (I am trying to do this one step at a time.)
 
  • #22
sophiecentaur said:
That is a fresh addition. Can you explain where it came from? (I am trying to do this one step at a time.)

the position function for a particle in SHM is x(t) = Xmcos(ωt+ø)

differentiating it you get velocity. and again you get a(t) = -ω[itex]^{2}[/itex]XmCos(ωt+ø) ... = a(t) = -ω[itex]^{2}[/itex]x(t)
 
  • #23
Right. So that tells you that ω2 equal to what? And hence ω= what?
 
  • #24
xtrubambinoxpr said:
-4x is the only one abiding by that principle
Correct. Nice job of figuring it out.

Chet
 
  • #25
sophiecentaur said:
Right. So that tells you that ω2 equal to what? And hence ω= what?

would it be Sqrt(a(t)/x(t))
 
  • #26
xtrubambinoxpr said:
would it be Sqrt(a(t)/x(t))
If, in general, a= -ω2x, and in your problem a = -4x, what is ω for your problem?
 
  • #27
Chestermiller said:
If, in general, a= -ω2x, and in your problem a = -4x, what is ω for your problem?

ω=2 because 2 squared = 4. so just to make sure I am doing -(w*w)(x)
I picked positive 5x before because I was an idiot and thought the negative was attached to the ω value.
 
  • #28
xtrubambinoxpr said:
ω=2 because 2 squared = 4. so just to make sure I am doing -(w*w)(x)
I picked positive 5x before because I was an idiot and thought the negative was attached to the ω value.
Don't be so hard on yourself. You just need to get a little more experience. Like I said earlier, you doped things out pretty well.

Chet
 
  • #29
xtrubambinoxpr said:
would it be Sqrt(a(t)/x(t))

That is not incorrect - but not what a sneaky teacher would do and doesn't actually give you the answer. It can be so annoying when you're shown the answer and you think "why did he choose to do it that way?". You just develop a feel for how to drag the right answer out of what you have. This time the way through is to substitute information in one equation with info from another - just spotting the similarity of two equations. Another tool for your maths toolbox.
 
  • #30
sophiecentaur said:
That is not incorrect - but not what a sneaky teacher would do and doesn't actually give you the answer. It can be so annoying when you're shown the answer and you think "why did he choose to do it that way?". You just develop a feel for how to drag the right answer out of what you have. This time the way through is to substitute information in one equation with info from another - just spotting the similarity of two equations. Another tool for your maths toolbox.
Thanks for the input! Appreciate it
 

1. What are oscillation problems?

Oscillation problems refer to any situation where a system or object moves back and forth repeatedly around a central point or equilibrium position. This can occur in various fields such as physics, engineering, and mathematics.

2. Why is it important to be able to solve oscillation problems?

Being able to solve oscillation problems is important because it allows us to understand and predict the behavior of systems and objects in motion. This knowledge can be applied in various real-world scenarios, such as designing stable structures or predicting the movement of celestial bodies.

3. What are some common techniques for solving oscillation problems?

Some common techniques for solving oscillation problems include using differential equations, applying Newton's laws of motion, and using energy conservation principles. These techniques can help us determine the amplitude, frequency, and period of oscillations.

4. How can I improve my problem-solving skills for oscillation problems?

To improve your problem-solving skills for oscillation problems, it is important to have a strong understanding of the underlying concepts and principles. Practice solving different types of oscillation problems and familiarize yourself with the various techniques and equations used. It can also be helpful to work with a study group or seek guidance from a teacher or tutor.

5. Are there any common mistakes to avoid when solving oscillation problems?

One common mistake to avoid when solving oscillation problems is using incorrect or outdated equations. It is important to ensure that the equations and principles being used are relevant to the specific problem being solved. It is also important to double-check calculations and units to avoid errors. Additionally, be sure to clearly define and label all variables and quantities in the problem.

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