SHM Equations: Oscillations: Edexcel A-level Physics

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  • Опубликовано: 25 авг 2024
  • All of the equations you need to be able to use for simple harmonic motion and some derivations so you know where they come from.
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Комментарии • 21

  • @emaan.1376
    @emaan.1376 Год назад +1

    thank you so much!

  • @maddybear8400
    @maddybear8400 Год назад

    At 11:00, when splitting down the weight of mass into components to get resultant force, why are we ignoring the components of tension? The horizontal component of tension is also acting towards equilibrium, right?

    • @nutshellphy
      @nutshellphy  Год назад +1

      Super question! If you look at what appears to be the horizontal component of tension though, it doesn’t point to the equilibrium position but rather to the centre of what would be a horizontal circle if this were being spun like that. As it’s oscillating, the centre of the circle is the top of that T line. So T provides the resultant force that keeps it in that vertical circle but not the force that pulls it along the arc. It is always perpendicular to the motion of the object at any instant. Does that make sense?

    • @maddybear8400
      @maddybear8400 Год назад +1

      @@nutshellphy Yeah, it did make a lot of sense to me! Thank you so much, sir, for your quick response!

  • @danaprppp8390
    @danaprppp8390 2 года назад

    so towards the equilibrium position it will always be mgsin theta for a pendulum and the weight is mgcos theta?

    • @nutshellphy
      @nutshellphy  2 года назад +1

      If theta is the angle at the top yes. They do like to mix that up though so be careful.

    • @danaabdulrida2248
      @danaabdulrida2248 2 года назад

      @@nutshellphy not always then (;_;)

    • @nutshellphy
      @nutshellphy  2 года назад +1

      There are only specific situations where it might not be… it would have to be fairly contrived but they like doing that for exam questions!

  • @tsmtrms53
    @tsmtrms53 Год назад

    excuse me sir , but in SHM maths the formula for the velocity, v, is the derivative of the formula of the displacement x, and when u find the derivative of cos it's going to - sin , which means the that the v is the gradient of the displacement graph and it will on make sense because if u look at the graph for the displacement and u draw a tangent to find the gradient of the graph , the gradeint will be negative , and the graph for the velocity starts from 0 and the amplitude is in the opposite side which makes it negative , so the formula for the velocity should be v = -A.w.sin(wt)
    maybe there is something that i didnt learn maybe like the formula that u showed us in the video applies only to a specific situation or something but i am only raising my concern to understand this topic even further, i have no intention of disrespecting u in any way

    • @nutshellphy
      @nutshellphy  Год назад

      You are absolutely correct! The equation is usually given as negative and of course, we should always think about the direction of the velocity relative to the displacement, as I pointed out in my video on the graphs. The reason I went down this route, instead of using differentiation, is because taking Maths or Further Maths is not a requirement for A-level physics and so not everyone would do it or be familiar with derivatives. This way, it uses concepts already covered (circular motion) to explain where these equations come from. These do not need to be derived for A-level physics and I did not want to introduce a whole mathematical idea (differentiation) in order for non-maths-taking people to get the idea. So this version of the derivation shows the magnitude of the velocity - a non-maths person can then think about the position and motion of the object to figure out its direction. Of course, it is easier if you are happy with differentiation because you don't have to do that!

    • @tsmtrms53
      @tsmtrms53 Год назад

      @@nutshellphy i understand that in most situations the negative indicates that its travelling in the opposite direction then ur positive direction , and from what i understood from you , after you derive the formula you will have to think about it logically while assuming which direction will be positive .
      so is this right ?

    • @nutshellphy
      @nutshellphy  Год назад

      The data sheet actually gives the equation with a negative so, in terms of calculating, one can go ahead and use that. You will never be asked to derive the formula at A-level. Many students, who have never studied sine curves, find these equations intimidating though so showing them where they come from can be helpful in dispelling some of the mystery around them. The derivation I did shows where it comes from, the sign attached to the equation just shows its direction.

    • @tsmtrms53
      @tsmtrms53 Год назад

      @@nutshellphy alright i understand , thank u so much!

  • @noor8361
    @noor8361 Год назад

    Can you do some question solving for Oscillations if possible before jan

    • @nutshellphy
      @nutshellphy  Год назад

      Sure… calculations or the written ones?

    • @noor8361
      @noor8361 Год назад

      @@nutshellphyfew questions from both, it would be really helpful!🙂

    • @nutshellphy
      @nutshellphy  Год назад +1

      Okay, I will get on it!

  • @danaprppp8390
    @danaprppp8390 2 года назад

    can u do a few questions on these if u can

    • @nutshellphy
      @nutshellphy  2 года назад

      I’ll see what I can dig out this weekend

  • @a.r1s
    @a.r1s Год назад

    Do we need to know the derivations of the equations or just how to use them?

    • @nutshellphy
      @nutshellphy  Год назад +1

      Just how to use them! The derivations were only for background info, to show connections with the rest of mechanics.