Simple Harmonic Motion Graphs IB Physics

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  • Опубликовано: 14 окт 2024

Комментарии • 25

  • @ZAsym
    @ZAsym 3 года назад +3

    thank you so SO much, I've been searching for HOURS for an explanation and this was by far the clearest and concise one. I seriously seriously appreciate this!

    • @danielm9463
      @danielm9463  3 года назад +1

      I'm so glad it helped! This is a tricky subject, and it REALLY helps me to think about a pendulum swinging back and forth or a mass-spring moving left and right. If you have any questions about the content, don't hesitate to post them here!

    • @ZAsym
      @ZAsym 3 года назад

      @@danielm9463 yup its easier to visualize, im back to revise cuz my exam's tomorrow!

  • @ironsugar5359
    @ironsugar5359 6 лет назад +17

    oh my gosh you're a life saver! Great vide

  • @dhaval-20
    @dhaval-20 6 лет назад +9

    Awesome presentations
    Helps me to Understand
    Oscillatory motion and S.h.m linked to it
    ☺😊

  • @1light4love
    @1light4love 4 года назад +3

    0:05 seconds in, he said "... like... THESE are motion graphs"... I was all in.
    "but its also true, for a PENDULUM!" 😅 and I have no idea why that was so exciting to me, but there I was alone in the kitchen shouting 'YAY PEN-DU-LUMS!!!!' 😂🤣😂
    your presence and energy are RADIANT my friend.
    YOU are gonna save the world. "
    with Simple Harmonic Motion.
    😊🙏💞

    • @danielm9463
      @danielm9463  4 года назад

      Thank you so much for such a kind message!! I say "YA PEN-DU-LUMS" on a regular basis.

  • @theJesai
    @theJesai 27 дней назад

    you're an AMAZING tutor. thanks so much :D

  • @msrlusted685
    @msrlusted685 6 лет назад +9

    Nice clear video thanks. Can you continue to show how to derive the equations for these graphs please?

    • @danielm9463
      @danielm9463  6 лет назад +2

      Thanks! Is there a specific equation you're thinking of?

    • @danielm9463
      @danielm9463  3 года назад

      It requires differential equations. SHM is defined as a system where the acceleration is proportional and opposite to the position x from equilibrium: a = -kx, where k is a constant. Acceleration a is the second time derivative of position x. So we have d²x/dt² = -kx. Intuitively, you may know from calculus class that the function we're looking for is sin(x) or cos(x). No other function has a second derivative that is proportional to the original function. For example, the first derivative of cos(x) is (d/dt)( cos[t] ) = -sin[t]. Taking the derivative of this gives the second derivative, (d/dt)(-sin[t]) = -cos[t]. I'm pretending that x = cos[t], though, which isn't quite right, because I've ignored the constant k. If you work it out, this constant ends up being ω², such that a = -ω²x and x = x₀cos(ωt). That's not a full derivation, because I'm skipping some steps, but it's hopefully enough to get you started so that you can search this question and find some good videos!

  • @rahmatpopal
    @rahmatpopal 6 лет назад +5

    You are an absolute legend!!

  • @tianiakukula8097
    @tianiakukula8097 Год назад

    Your surprise at the block motion at velocity makes me suprised as well 😂

  • @saqibshams5143
    @saqibshams5143 5 лет назад +4

    Best explanation ever!!

  • @boboganbobogan9297
    @boboganbobogan9297 2 года назад +1

    Thank you so much, it helped me a lot. I have a question, when a block is at equilibrium position acceleration is zero but velocity is a maximum value but how is this possible when there is velocity and no acceleration?

    • @danielm9463
      @danielm9463  2 года назад +1

      The velocity is a result of its previous acceleration. The block accelerates from the edge all the way till it reaches the middle. That acceleration speeds it up, and briefly (while at the middle) the block is just gliding/coasting at a high speed due to its inertia. It's like being in a car on the highway, speeding up, and then coasting for a brief moment before hitting the brakes. The inertia of the block carries it through the center position, and as soon as it passes to the other side of the center/equilibrium position, it begins slowing down.

  • @sarujan02thangavel93
    @sarujan02thangavel93 3 года назад +1

    Thank you very much sir

  • @farisalsaleh411
    @farisalsaleh411 5 лет назад +3

    Thank you so much!!

  • @thulithainduwara5065
    @thulithainduwara5065 3 года назад

    It was really really helped me. Thank you so much 👍😊

  • @tahsintasnim9440
    @tahsintasnim9440 5 лет назад +2

    Alrighty then
    Thank you for the video

  • @makgokacharmaine833
    @makgokacharmaine833 5 лет назад +1

    I'm writing in 1 hour and I must say thank you so much