How can Simple Harmonic Motion have angular frequency?

  1. It isn't making any intuitive sense. If it isn't moving in circular motion, how can it have angular frequency or speed?

    Also,

    [tex]v=\pm ω\sqrt { A^{ 2 }-x^{ 2 } }[/tex]

    only applies to SHM with springs only, right?

    Also, does anyone know how to derive this equation below?

    [tex]x=\frac { \pm \sqrt { { { v }_{ max } }^{ 2 }+{ v }^{ 2 } } }{ ω }[/tex]
     
    Last edited: Feb 12, 2013
  2. jcsd
  3. sophiecentaur

    sophiecentaur 13,700
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    Gold Member

    Chegg
    The (differential) equation of motion for a particle of mass m on a spring with spring constant k and a displacement x from the equilibrium position is
    m d2x/dt2 = -kx
    When you solve this, you find a solution of the form
    x= A sin(ωt - ∅)
    ∅ is an arbitrary value for the phase of the oscillation.
    That's where the variable ω comes from. Trig functions involve angles so ω has the dimension of an angle divided by time. Hence it's referred to as an angular frequency. There is one other point and that is that ω is in radians (as all good angles are) so is 2∏f, where f is the frequency in cycles per second (Hz).
     
  4. That doesn't really help me intuitively to see where the angular speed comes from.

    Also, since you solved the differential equation of motion for a particle with a spring, does that mean that the solution only applies to simple harmonic motion with springs? Is there simple harmonic motion without a spring?
     
  5. sophiecentaur

    sophiecentaur 13,700
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    You can't always expect to have the luxury of that, I'm afraid. If you have ever solved a second order diff equation, you will know that the result is in the form of a trig function. You just have to accept that.

    It doesn't have to be a spring. That equation of motion applies to many physical situations - where a 'restoring force' happens to be proportional to displacement from an equilibrium position. A simple pendulum approximates to this for small displacements from the vertical, too. Why not look it up?????

    BTW, the Negative sign in the equation is because the direction of the restoring force is opposite to the direction in which the particle has been displaced (i.e. back where it came from). If you are not familiar with equations of motion then I am sure you will come across them before too long and things may get clearer.
     
  6. So when the pendulum is coming back to equilibrium, it becomes positive because the particle is being displaced in the same direction as the restoring force?
     
  7. sophiecentaur

    sophiecentaur 13,700
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    Gold Member

    When the position is the other side of the origin (i.e. the displacement is Negative), the force becomes positive. The situation is symmetrical for both sides, the force is always back to the equilibrium position. Draw it out for both cases.
     
  8. I think I get why it's negative now.

    By convention, we assign the right direction as the positive direction.

    Let's take a look at the term mgsin(θ). Assuming m and g are always positive, the term changes sign depending on what θ is.

    When θ is positive, sin(θ) is positive and the term mgsin(θ) is positive. But we know the restoring force is to the left when θ is positive, therefore, the force must be negative and so we add a negative sign.

    When θ is negative, sin(θ) is negative and the term mgsin(θ) becomes negative. But we know that the restoring force is to the right when θ is negative, therefore, the force must be positive, and so we add a negative sign.

    It works even when you assign the left as the positive direction.

    Part of why I didn't understand you was because I didn't actually understand displacement. Displacement is position away from the equilibrium. It is positive when the bob is to the right of the equilibrium and left when the bob is to the left of the equilibrium. For some reason, I thought displacement was instantaneous velocity.
     
  9. I've realized that a lot of simple harmonic motion is basically periodic motion in which the restoring force is proportional to the position away from the equilibrium. Does that mean all simple harmonic motion can be modelled using a spring?

    The equation:

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

    applies to simple harmonic motion when a spring is involved where k is the spring constant. But k doesn't need to be the spring constant right? It just needs to be the proportional constant of the restoring force to the displacement, right?

    [tex]k=\frac { F }{ x }[/tex]

    where F is any restoring force.

    Now my question is: is this ever useful when dealing with simple harmonic motion problem where the question doesn't involve a spring? The issue is that problems don't give you the proportional constant of the restoring force to the displacement when there is no spring involved I'm assuming?

    [tex]θ(t)={ θ }_{ max }cos(ωt+ϕ)[/tex]

    Does θmax need to be in radians or degrees? It doesn't matter, does it? Will it matter if I take the derivative?

    [tex]w(t)={ -θ }_{ max }ωsin(ωt+ϕ)[/tex]

    How can angular speed be a function of itself?
     
    Last edited: Feb 12, 2013
  10. WannabeNewton

    WannabeNewton 5,775
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    The general solution to the DE [itex]r'' + \omega ^{2}r = 0[/itex] would be [itex]r(t) = Ae^{i\omega t} + Be^{-i\omega t} [/itex]. For simplicity of visualization, take A = 1, B = 0 so that [itex]r(t) = e^{i\omega t}[/itex]. You will recognize this as being [itex]exp:I\rightarrow S^{1}[/itex] where [itex]S^{1}[/itex] is the unit circle (we are viewing it as being a subset of the complex plane) and [itex]I[/itex] is an appropriate interval in [itex]\mathbb{R}[/itex]. Now, with regards to the physics of simple harmonic motion, we just take the real part of the above expression for [itex]r(t)[/itex] but you can visualize [itex]\omega [/itex], the angular frequency, as being the rate at which the unit circle is being swept out in the complex plane.
     
  11. WannabeNewton

    WannabeNewton 5,775
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    Wildly, wildly, wildly, wildly useful. SHM will show up over and over and over again in physics in various forms other than actual, physical springs. A couple of specific examples are: simple pendulum motion, a positive charge given a small kick at the center of a ring of uniform positive charge density will undergo simple harmonic motion, after some idealizations if you drill a hole in the earth and drop down you will execute SHM, and a marble rolling in a circular dish in the small angle approximation will undergo SHM. There are so many rich examples of SHM out there; you will find a plethora of examples on the internet. By showing the object undergoes SHM through newton's 2nd law you can read off the proportionality constant from the DE.
     
  12. Thanks for the insight sophiecentaur and WannabeNewton.

    FYI, the only questions left are:

    and

     
  13. WannabeNewton

    WannabeNewton 5,775
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    Are you sure it is [itex]v_{max}^{2} + v^{2}[/itex] as opposed to [itex]v_{max}^{2} - v^{2}[/itex]? Otherwise you get that when [itex]v = v_{max}[/itex], [itex]\omega x = \pm \sqrt{2v_{max}^{2}}[/itex] but [itex]x(v = v_{max})[/itex] should be zero since the maximum speed will occur at the equilibrium position (I am assuming this is for simple harmonic motion and that [itex]x[/itex] is the displacement from equilibrium as usual).
     
  14. Opps. Error on my part. Don't worry too much about the derivation. The second question is more important for me at the moment.
     
  15. WannabeNewton

    WannabeNewton 5,775
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    I would like you to derive it yourself. It is very simple; just use the fact that this system conserves the total energy and that, by definition, ω2 = k / m. As for the second question, you are confusing the angular frequency of the period of the motion with the angular velocity of the mass. It is unfortunate that both have the same symbol for the particular case you are working with but use [itex]\dot{\theta }[/itex] instead for the angular velocity.
     
  16. sophiecentaur

    sophiecentaur 13,700
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    I already told you that all angles should be in radians - or you keep getting 2pi all over the place. For a start - your derivative is only correct if ω is in radians. Look up differentiating trig functions - basics of. Degrees suck in Science. Radians will be used all over the Universe (literally) by any civilisation with Maths but degrees are totally arbitrary.

    I don't know what those equations mean as you didn't define the variables so it is not possible to derive them.
    The "w" on the left hand side is not omega! And what is the theta? Is this describing a tortional vibration? Those angles aren't the same angles as the angular velocity in the vibration cycle. See how confusing it is when the variables aren't defined.
     
  17. I understand that the argument of a trigonometric function needs to be in radians (because the derivative of sinθ is only cosθ when θ is in radians), but what if it's outside the argument as it is in the equation of simple harmonic motion for a simple pendulum?

    [tex]θ(t)={ θ }_{ max }cos(ωt+ϕ)[/tex]

    Opps, the "w" should've been omega. I took the derivative of the equation of simple harmonic motion for a simple pendulum. θ is the angle between the vertical line pointing towards the ground and the string of the pendulum.

    I accept the challenge.

    Oh. That's quite confusing to the novice. Makes sense now.
     
  18. jtbell

    Staff: Mentor

    See the section of this video from about 1:45 to 2:07. It demonstrates the connection between SHM and uniform circular motion.

     
    Last edited by a moderator: Sep 25, 2014
  19. Derived it!

    [tex]TE={ KE }+PE\quad and\quad TE={ KE }_{ max }={ PE }_{ max }\\ \\ { KE }_{ max }=KE+PE\\ \frac { 1 }{ 2 } m{ { v }_{ max } }^{ 2 }=\frac { 1 }{ 2 } m{ { v } }^{ 2 }+\frac { 1 }{ 2 } k{ x }^{ 2 }\\ m{ { v }_{ max } }^{ 2 }=m{ { v } }^{ 2 }+k{ x }^{ 2 }\\ k{ x }^{ 2 }=m{ { v }_{ max } }^{ 2 }-m{ { v } }^{ 2 }\\ { x }^{ 2 }=\frac { m }{ k } ({ { v }_{ max } }^{ 2 }-{ { v } }^{ 2 })\\ { x }^{ 2 }={ ω }^{ -2 }({ { v }_{ max } }^{ 2 }-{ { v } }^{ 2 })\\ x=\pm \sqrt { { ω }^{ -2 }({ { v }_{ max } }^{ 2 }-{ { v } }^{ 2 }) } \\ x=\pm \frac { \sqrt { { { v }_{ max } }^{ 2 }-{ { v } }^{ 2 } } }{ { ω } } [/tex]
     
    Last edited: Feb 12, 2013
  20. WannabeNewton

    WannabeNewton 5,775
    Science Advisor

    Nice!
     
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