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    Hi,

    Say I have a particle performing SHM between two points A and B

    A------------M---------C----B

    Where M is the middle, would the time it takes to get from C to B = to the time it takes to get from B to C? Will this also apply to any two points you pick?

    Thnaks
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    Yes.
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    (Original post by 2710)
    Hi,

    Say I have a particle performing SHM between two points A and B

    A------------M---------C----B

    Where M is the middle, would the time it takes to get from C to B = to the time it takes to get from B to C? Will this also apply to any two points you pick?

    Thnaks
    Yes, time from C to B = time from B to C. If you sketch a displacement-time graph, you will see it forms a sinusoidal graph, and due to the symmetry of this graph, the time to get from any point P to any point Q will be the same as the time to get from Q to P (this is assuming that when the point is at P, if it going towards Q, then in the other situation we need the particle to be going towards P when it is at Q [the same is true for when the particle is going away from Q when at P]).
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    OK can you please tell me where I have gone wrong then:

    A-------P---M-------Q----B

    A to B is 8m
    Period of Osc = 12s
    P and Q are 3 and 6 m from A respectively.

     \omega = \frac {\pi}{6}

    Find the time it takes to get from P to Q. The MS uses the x = A sinwt formula. But I want to use the cos one as a challenge.

    x = A cos wt

    P : 3 = 4cos (\frac {\pi}{6} t )
    t = 1.38

    For Q, I tried using x as 6, but then I cant cos inverse 6/4, am I doing something wrong? So I just did the other side, and then minused:

    Q: 2= 4 cos(\frac {\pi}{6} t )
    t = 2 therefore t = 6 - 2 = 4

     4 - 1.38 = 2.62

    Which is the wrong answer.

    Any help appreciated
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    (Original post by 2710)
    OK can you please tell me where I have gone wrong then:

    A-------P---M-------Q----B

    A to B is 8m
    Period of Osc = 12s
    P and Q are 3 and 6 m from A respectively.

     \omega = \frac {\pi}{6}

    Find the time it takes to get from P to Q. The MS uses the x = A sinwt formula. But I want to use the cos one as a challenge.

    x = A cos wt

    P : 3 = 4cos (\frac {\pi}{6} t )
    t = 1.38

    For Q, I tried using x as 6, but then I cant cos inverse 6/4, am I doing something wrong? So I just did the other side, and then minused:

    Q: 2= 4 cos(\frac {\pi}{6} t )
    t = 2 therefore t = 6 - 2 = 4

     4 - 1.38 = 2.62

    Which is the wrong answer.

    Any help appreciated
    the displacement of P isn't 3, its 1 right?

    coz x is the displacement from M
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    (Original post by 2710)
    OK can you please tell me where I have gone wrong then:

    A-------P---M-------Q----B

    A to B is 8m
    Period of Osc = 12s
    P and Q are 3 and 6 m from A respectively.

     \omega = \frac {\pi}{6}

    Find the time it takes to get from P to Q. The MS uses the x = A sinwt formula. But I want to use the cos one as a challenge.

    x = A cos wt

    P : 3 = 4cos (\frac {\pi}{6} t )
    t = 1.38

    For Q, I tried using x as 6, but then I cant cos inverse 6/4, am I doing something wrong? So I just did the other side, and then minused:

    Q: 2= 4 cos(\frac {\pi}{6} t )
    t = 2 therefore t = 6 - 2 = 4

     4 - 1.38 = 2.62

    Which is the wrong answer.

    Any help appreciated
    P : 1 = 4cos (\frac {\pi}{6} t )

    Q: 2= 4 cos(\frac {\pi}{6} t )

    then minus them. try that and see?
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    (Original post by peaches6994)
    P : 1 = 4cos (\frac {\pi}{6} t )

    Q: 2= 4 cos(\frac {\pi}{6} t )

    then minus them. try that and see?
    No no no! Im using the cos formula, not the sine one. "Cos" you take distance form the end, "sin" you take from the centre. The way you did it is the way the MS did, which is what I dont want to do. Besides, its -1.

    So can anyone see the flaw in my method?
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    are you sure? I think x is fixed, no matter what formula you use.
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    (Original post by 2710)
    No no no! Im using the cos formula, not the sine one. "Cos" you take distance form the end, "sin" you take from the centre. The way you did it is the way the MS did, which is what I dont want to do. Besides, its -1.

    So can anyone see the flaw in my method?
    are you familiar with the considering the SHm as a circular motion and finding the time , I find it much more logical and easy to follow

    Also Q is 2m away from M so I think x should be 2. so time taken to travel from Q to M ,2=4coswt
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    the answer should be 5.38 given you calculations are correct
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    (Original post by 2710)
    OK can you please tell me where I have gone wrong then:

    A-------P---M-------Q----B

    A to B is 8m
    Period of Osc = 12s
    P and Q are 3 and 6 m from A respectively.

     \omega = \frac {\pi}{6}

    Find the time it takes to get from P to Q. The MS uses the x = A sinwt formula. But I want to use the cos one as a challenge.

    x = A cos wt

    P : 3 = 4cos (\frac {\pi}{6} t )
    t = 1.38

    For Q, I tried using x as 6, but then I cant cos inverse 6/4, am I doing something wrong? So I just did the other side, and then minused:

    Q: 2= 4 cos(\frac {\pi}{6} t )
    t = 2 therefore t = 6 - 2 = 4

     4 - 1.38 = 2.62

    Which is the wrong answer.

    Any help appreciated
    In the last step you should add 4+1.38 as 4 is the time you take to get from M to Q and 1.38 the time to get from P to M.
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    (Original post by rbnphlp)
    are you familiar with the considering the SHm as a circular motion and finding the time , I find it much more logical and easy to follow

    Also Q is 2m away from M so I think x should be 2. so time taken to travel from Q to M ,2=4coswt

    OP .... the above is an important post and I would suggest that you consider the method rbnphlp suggests.
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    (Original post by 2710)
    No no no! Im using the cos formula, not the sine one. "Cos" you take distance form the end, "sin" you take from the centre.
    No. For SHM, the distance is always from the center (of the harmonic motion).
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    Ah, so I tkae distance from the centre, BUT time is the one which varies right?

    With the cos formula I take the time from the end. Ok, Let me try that.

    (Original post by ananda2010)
    the answer should be 5.38 given you calculations are correct
    Sorry, but thats wrong also :P
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    First of all, thanks to everyone who helped. I appreciate your help with other solutions, but I just wanted to find the flaw in my working using my method. And yes, I am familiar with the circular motion thing, but I just wanted to use this method

    OK, I've seem to have got my answer, but I'm a tad confused:

    -----3m-----1m]-----2m-----2m
    A----------P----M-------Q-------B

    P : -1 = 4cos (\frac {\pi}{6} t )
    t = 3.48

    But this is the weird part. I am taking my time from A. And I put into the formula x=-1. The time it gives me is over 3 seconds (the time it takes to get to the middle), when it is clear that the time it takes to get to P should be less than 3. Ie, the formula takes negative distances to be on the other side of where you take your time to be. Is this meant to be? I always though negative meant (in this case) left of M.
    Anyways:

    Q: 2= 4 cos(\frac {\pi}{6} t )
    t = 2

    Same here, ie, positive x =2 and it takes it to be on the left hand side of M, ie 2m from M

    So basically, my formula flips my whole diagram right to left..kinda >__> which I found hard to get in my head.

    Besides that 3.48-2 = 1.48, which is the correct answer.

    I just need someone to confirm my thinking for x distances, I hope you understand what I mean

    THANKS help appreciated

    EDIT: Ah, I know an easier way to help you understand me.

    For example, say the question asked me the time it took to get from A to P.

    So I use this:

    P : -1 = 4cos (\frac {\pi}{6} t )
    t = 3.48

    Which is obviously the wrong answer. I have to do 3.48 - 3, since it is on the wrong side. Do you understand what I'm getting at?
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    Bump ? T__T
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    Anyone ? I really need to know this
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    (Original post by 2710)
    For example, say the question asked me the time it took to get from A to P.

    So I use this:

    P : -1 = 4cos (\frac {\pi}{6} t )
    t = 3.48

    Which is obviously the wrong answer. I have to do 3.48 - 3, since it is on the wrong side. Do you understand what I'm getting at?
    Notice that using the cosine, at time t=0 the displacement is 4, which is the position of A relative to the centre. Note: it is positive.

    The position of P will be 1, not -1, as it is 3 units away from A.

    Edit: Whichever end you have chosen as the starting point (if you are using the cosine) determines the positive direction.
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    (Original post by ghostwalker)
    Notice that using the cosine, at time t=0 the displacement is 4, which is the position of A relative to the centre. Note: it is positive.

    The position of P will be 1, not -1, as it is 3 units away from A.

    Edit: Whichever end you have chosen as the starting point (if you are using the cosine) determines the positive direction.

    Ok let me think about this. So if I had to find A to Q, do I put x as negative 2?

    Thanks

    EDIT: Nevermind it works! Thanks ghostwalker
 
 
 
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