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B Angular Velocity in Simple Harmonic Motion

  1. Nov 2, 2017 #1

    Abu

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    I am very confused about angular velocity ω and why its used in simple harmonic motion. ω is described as θ/τ but when it comes to masses on springs, there is no angle - it is zero. Angular velocity comes from circular motion but the motion of SHM is not circular. My confusion is even greater when finding that in order to prove acceleration is directly proportional to displacement the equation is -ω^2x. I looked at how to prove this formula but couldn't understand it as I have no knowledge of calculus and derivatives.

    Anyways, I am just confused on why angular velocity is used in simple harmonic motion and how -ω^2x is representative of displacement.

    If my question does not make sense just let me know I'll be more than happy to clarify it.

    Thank you.
     
  2. jcsd
  3. Nov 2, 2017 #2

    vanhees71

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    If you have a mass on a spring, the equation of motion is
    $$m \ddot{x}=-k x.$$
    What are the solutions of this differential equations? Try to solve it yourself first, before we provide the answer!
     
  4. Nov 2, 2017 #3
    Simple harmonic motion is a "projection" of circular motion. I suggest the following video to ubderstand the concept.
     
  5. Nov 2, 2017 #4

    Mister T

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    Did you notice that the OP declared no knowledge of calculus?
     
  6. Nov 2, 2017 #5

    Mister T

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    ##\omega x^2## is the acceleration, not the displacement.

    ##\omega=2 \pi f## where ##f## is the frequency of the oscillator. And if ##\omega## is the rotation rate of a particle executing uniform circular motion, then it's shadow, as shown in Post #3, executes simple harmonic motion with frequency ##f##. That is the connection.
     
  7. Nov 2, 2017 #6

    Abu

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    Thank you everyone for your replies, it cleared up most of my confusion. Sorry for the late response as well. The video that Chandra posted lead me on a string of other videos that explained the similarities between circular motion and the motion of a mass on a spring. I still don't understand how to derive the formula acceleration = -w^2 x, but that is because it involves calculus and I am assuming that's just something I have to learn to accept until I learn about derivatives and such.

    But from my understanding after watching all of these videos and reading your replies, and feel free to correct me if I am wrong, is that:
    • The formula acceleration = -w^2 x proves that acceleration is proportional to displacement because x represents displacement (something that I completely did not realize earlier)
    • Angular velocity ω is used for simple harmonic motion because ω = θ/t, which is 2π/τ because one revolution is 360 degrees and T is the period.. This can be inverted to make the formula Mister T mentioned which is 2πƒ.
    • The shadow shows how at the two end points of the circles rotation the velocity is equal to zero, whereas in the middle it is at its maximum, similar to a mass on a spring
    I still need to practice to reinforce my understanding of these concepts, but thank you everyone for your input, help and criticism. Feel free to correct or add to anything I said.
     
  8. Nov 2, 2017 #7

    Mister T

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    It can be done without calculus. Look in any trigonometry-based college-level introductory physics textbook. These textbooks are very common as they are used for the course taken by pre-meds.
     
  9. Nov 2, 2017 #8

    Abu

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    This isn't trigonometry, but I think this would work too:

    Could I say that the acceleration is simply centripetal acceleration because the motion is circular? And that if it is centripetal acceleration the formula is
    a = V^2/r
    And we know that linear velocity is equal to: v = wr
    Putting that into a = V^2/r we get a = (wr)^2/r
    this would make a = w^2r and I am assuming that r in this formula is the same as x in the acceleration = -w^2 x
    Can you tell me if this is right? The only thing that I notice is different is that the negative sign for w is not prevalent in a = w^2r

    Thank you.
     
  10. Nov 3, 2017 #9

    vanhees71

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    No, sorry. But then he or she has no chance to understand any physics. The good news is that calculus is an interesting subject in itself and you can have a lot of fun in learning it!
     
  11. Nov 3, 2017 #10

    Mister T

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    It depends on what you're trying to do. The minus sign arises when you look at the components of the displacement, velocity, and acceleration rather than their magnitudes. This is what is done when you study one-dimensional motion although it's seldom stated explicitly.

    The radius of the circular motion equals the amplitude (largest value of displacement) of the simple harmonic motion.

    All of this is explained quite thoroughly.
     
  12. Nov 3, 2017 #11

    Mister T

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    Learners can build up their knowledge of both physics and calculus concepts together. This is the way most people learn it, and it's historically the way the knowledge structure was created.
     
  13. Nov 3, 2017 #12

    vanhees71

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    Yes, I agree. Physics is a great way to get intuition about math, and it should be taught in this way, but it should be taught in this way! I was shocked when I learnt some years ago when being a postdoc in the US that there are textbooks today called "Calculus free Physics". I once had to substitute for a colleague in giving some lessons in his calculus free physics lecture. This was the most difficult task to make sense of this oxymoron in trying to explain Newtonian mechanics. As you say, it's no surprise that Newton was among the first discoverers of calculus (although I'm clearly a Leibnizian).
     
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