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    I am having confusion with this Oscillation related displacement Equation mentioned in my syllabus:

    X = X0 sin ωt

    What does X0 stand for ?


    I checked my text book but there was a completely different equation for displacement:

    X = r cos ωt


    Where r is the amplitude of the oscillation (radius of circle).

    This one is understandable as it is proved by comparing Oscillations with Circular motion.

    Since x = r cos θ and θ = ωt

    But I have to know about the one in my syllabus not the one in the text book.

    There is the similar problem with the velocity equation:

    In the syllabus:


    V = V0 cos ωt

    { what is V0 }



    In the text book:



    V = - ωr sin ωt

    I am really confused …
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    Xo is initial displacement.
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    basically the 0 signifies t=0.
    So when t=0, x= x0 and v=v0,
    which are initial displacement and velocity
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    if you draw a right angled triangle with x0 and r as the shorter sides, you'll see that rcosA=x0sinA where A is any angle.
    Which is why both equations can be used.
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    and of course, sin and cos are identical functions but with different starting points.
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    (Original post by DominicJ)
    Xo is initial displacement.
    It can't be initial displacement. If it was, then the particle (whatever you want to call it) couldn't ever go beyond where it started.

    Start it in the centre, even with a speed, and by using X0 as the initial displacement, it can't move anywhere with that forumla.

    Oh, and OP, Sin is to find the displacement when it starts at the centre. Cos is to find the displacement when it starts at an end. The textbook was probably building up to starting it from an end, turn over a few pages and it should have the cos version.
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    Thanks a lot
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    (Original post by Charlybob)
    Oh, and OP, Sin is to find the displacement when it starts at the centre. Cos is to find the displacement when it starts at an end. The textbook was probably building up to starting it from an end, turn over a few pages and it should have the cos version.
    The book doe not have a cos version, I checked.
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    (Original post by Hashmi)
    I am having confusion with this Oscillation related displacement Equation mentioned in my syllabus:

    X = X0 sin ωt

    What does X0 stand for ?
    X0 is the maximum displacement of the particle from the centre of the oscillations, usually known as the amplitude. I've done this in maths, and it's always referred to as "a" where I've seen it, or it could also be "r" as you've put below. Regardless of what letter it is, it's referring to the amplitude.

    I checked my text book but there was a completely different equation for displacement:

    X = r cos ωt
    The two equations that you've posted so far are both correct, however, it's measuring "t" from a different "starting point". (I'm going to call the amplitude "r" in the following explanation, although it could be "X0" or "a".)

    The general equation for SHM is
    x = r \sin (\omega t + \alpha)
    where \alpha is a constant. Now, if t = 0 when x = 0, then \sin \alpha = 0, or, \alpha = 0 and hence the equation for the SHM is x = r \sin (\omega t), as was stated in the first equation you gave in your post.

    Now consider the case where x = r when t = 0 (that is, the particle is initally at the point of maximum displacement). Then you get
    \sin \alpha = 1 which, if you're working in radians (which you should be) gives you  \alpha = \frac{\pi}{2}.
    So you get x = r \sin (\omega t + \frac{\pi}{2}). However, you should notice that \sin (\theta + \frac{\pi}{2}) = \cos \theta, which is always true (think about the graphs of sin and cos), and hence in this case the simplest equation for the SHM is
    x = r \cos (\omega t). If the particle is at some other point when t = 0, then you'll get a different value of alpha which you'll have to calculate.

    To summarise, if the particle is initially at the centre of the motion, you use sin, and if it's initally at the maximum displacement, you use cos.



    There is the similar problem with the velocity equation:

    In the syllabus:


    V = V0 cos ωt


    { what is V0 }



    In the text book:



    V = - ωr sin ωt

    I am really confused …
    V0 means the maximum velocity. The textbook and syllabus have explained it poorly, arbitrarily switching between sin and cos, but it could be either, depending on whether it's at maximum or zero velocity when t = 0.

    Also, V0 = -wr is true in this case. If you've done A-level maths (C3/C4, depending on your exam board), you can show this using differentiation of the expression of the displacement.
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    (Original post by Charlybob)
    Oh, and OP, Sin is to find the displacement when it starts at the centre. Cos is to find the displacement when it starts at an end. The textbook was probably building up to starting it from an end, turn over a few pages and it should have the cos version.
    The book doe not have a cos version, I checked.

    So ... from what you are saying, the equations are same, so V0 is ωr X0 is r. The only difference being point of start.


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    My bad.
    x0 is max displacement from centre point.
    Generally i've used this with springs where max displacement=initial displacement.
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    (Original post by tommm)
    X0 is the maximum displacement of the particle from the centre of the oscillations, usually known as the amplitude. I've done this in maths, and it's always referred to as "a" where I've seen it, or it could also be "r" as you've put below. Regardless of what letter it is, it's referring to the amplitude.



    The two equations that you've posted so far are both correct, however, it's measuring "t" from a different "starting point". (I'm going to call the amplitude "r" in the following explanation, although it could be "X0" or "a".)

    The general equation for SHM is

    where is a constant. Now, if t = 0 when x = 0, then , or, and hence the equation for the SHM is , as was stated in the first equation you gave in your post.

    Now consider the case where x = r when t = 0 (that is, the particle is initally at the point of maximum displacement). Then you get
    which, if you're working in radians (which you should be) gives you .
    So you get . However, you should notice that , which is always true (think about the graphs of sin and cos), and hence in this case the simplest equation for the SHM is
    . If the particle is at some other point when t = 0, then you'll get a different value of alpha which you'll have to calculate.

    To summarise, if the particle is initially at the centre of the motion, you use sin, and if it's initally at the maximum displacement, you use cos.





    V0 means the maximum velocity. The textbook and syllabus have explained it poorly, arbitrarily switching between sin and cos, but it could be either, depending on whether it's at maximum or zero velocity when t = 0.

    Also, V0 = -wr is true in this case. If you've done A-level maths (C3/C4, depending on your exam board), you can show this using differentiation of the expression of the displacement.
    Ah, thanks its quite clear now
 
 
 
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