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How to work out K?

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    my problem is: that the equation is:

    A= ae^(-kn)

    a is the original amplitude which is 0.1m
    n= number of swings.

    k= what I/we need to find.

    Now when I done the experiment It took 30 swings to get to 0.049m.

    But the problem is I have to put this on the graph

    so for example after the first swing the value of A= 0.097m.
    and it decreased so on, up to the 30th swing to 0.049m.

    each swing is 2 seconds long.

    Now this is how I plotted the graph, I plotted the graph as decrease in A over time as 30 swings= 60 seconds.

    But does that mean the gradient from this line will not give me the value of k?

    Because was I suppose to draw a log log graph.

    If i was supposed to draw a log log graph how am I meant to do it?
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    How many threads on the same question

    Is this coursework of some kind?
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    (Original post by TenOfThem)
    How many threads on the same question

    Is this coursework of some kind?
    Well in two weeks time I need to do coursework but I can't even complete this guide work so I will be dead in two weeks.
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    (Original post by Freddy12345)
    my problem is: that the equation is:

    A= ae^(-kn)

    a is the original amplitude which is 0.1m
    n= number of swings.

    k= what I/we need to find.

    Now when I done the experiment It took 30 swings to get to 0.049m.

    But the problem is I have to put this on the graph

    so for example after the first swing the value of A= 0.097m.
    and it decreased so on, up to the 30th swing to 0.049m.

    each swing is 2 seconds long.

    Now this is how I plotted the graph, I plotted the graph as decrease in A over time as 30 swings= 60 seconds.

    But does that mean the gradient from this line will not give me the value of k?

    Because was I suppose to draw a log log graph.

    If i was supposed to draw a log log graph how am I meant to do it?
    You're not supposed to draw a loglog graph. Just a log graph.

    Note that given A=ae^(-kn) taking logs gives up logA=loga-kn. If we think of logA as a new variable (call it y) then we have

    y=-kn+loga (look familiar?) so this is just the equation of a line in the plane. the gradient is then obviously going to match up with -k.
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    (Original post by Totally Tom)
    You're not supposed to draw a loglog graph. Just a log graph.

    Note that given A=ae^(-kn) taking logs gives up logA=loga-kn. If we think of logA as a new variable (call it y) then we have

    y=-kn+loga (look familiar?) so this is just the equation of a line in the plane. the gradient is then obviously going to match up with -k.
    I still don't understand. isn't it mean to be ln than log?
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    (Original post by Freddy12345)
    I still don't understand. isn't it mean to be ln than log?
    nobody in the real world uses ln. every log is base e.

    is there anything else other than the notation that you don't understand?
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    I just don't understand. I Know you guys are making it as simple as possible but I just don't get it!
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    (Original post by Freddy12345)
    I just don't understand. I Know you guys are making it as simple as possible but I just don't get it!
    then perhaps you need to talk to your teacher
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    (Original post by TenOfThem)
    then perhaps you need to talk to your teacher

    I sent you a message.
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    (Original post by Freddy12345)
    I just don't understand. I Know you guys are making it as simple as possible but I just don't get it!
    We can't help you if you don't tell us what you don't understand.

    Repeating "I don't understand" doesn't tell us what you don't understand.
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    (Original post by Freddy12345)
    I sent you a message.
    I saw

    I will repeat : talk to your teacher
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    (Original post by Totally Tom)
    We can't help you if you don't tell us what you don't understand.

    Repeating "I don't understand" doesn't tell us what you don't understand.
    What I don't understand is, is when i plot the graph will i get a curved line or a straight line because initially this was the graph that I did:

    Displacement against time, in which at the peak of displacement is the amplitude so I only care about the peaks.

    Therefore my graph is now amplitude over time but as in without any logs, just length and time.

    The length decreases over time, so I get sort of a curved decrease, like an exponential decay but a weak one. So what I am trying to say is. Can I just pick a point of the graph sub into the equation and find out the k that way or is my graph wrong in the first place.
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    Ok let's try to make it even easier.

    You know a = 0.1 so your equation is A=0.1 e^{-kn}

    Multiply both sides of the equation by 10.

    10A = e^{-kn}

    Take the natural logarithm of each side of the equation.

    \ln (10A) = -kn

    Now if you plot \ln (10A) on the y axis and n on the x axis, the points should form a straight line with a negative gradient.

    Draw a line of best fit and calculate the gradient. The magnitude of the gradient is k, the number you are looking for.
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    (Original post by mr m)
    ok let's try to make it even easier.

    You know a = 0.1 so your equation is a=0.1 e^{-kn}

    multiply both sides of the equation by 10.

    10a = e^{-kn}

    take the natural logarithm of each side of the equation.

    \ln (10a) = -kn

    now if you plot \ln (10a) on the y axis and n on the x axis, the points should form a straight line with a negative gradient.

    Draw a line of best fit and calculate the gradient. The magnitude of the gradient is k, the number you are looking for.
    thank you!
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    Aslo thanks to the other two who helped me as well.
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    (Original post by Freddy12345)
    thank you!
    You should make sure your straight line passes through (0, 0) by the way. Please note that this is NOT a general rule - straight lines do not always have to pass through (0, 0).

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Updated: April 13, 2012
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