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    I have the answer to this. It's 25 N. But I can't get to that answer at all.

    Here is the question and my workings on the attachment.

    What should I try now? Thanks.
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    (Original post by maggiehodgson)
    I have the answer to this. It's 25 N. But I can't get to that answer at all.

    Here is the question and my workings on the attachment.

    What should I try now? Thanks.
    I also got 25

    there is a numerical error somewhere
    will now look for it
    last line you are missing the 400
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    (Original post by maggiehodgson)
    I have the answer to this. It's 25 N. But I can't get to that answer at all.

    Here is the question and my workings on the attachment.

    What should I try now? Thanks.
    On the penultimate line, you have: 400(0.25g + \lambda x) - 100g= -100\ddot{x}, you should then expand this to get: 400 \lambda x = -100\ddot{x}, you've left it as \lambda x

    Sorry if the LaTeX doesn't work, I'm on my phone and it's very hard to type. :lol:
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    (Original post by TeeEm)
    I also got 25

    there is a numerical error somewhere
    will now look for it
    last line you are missing the 400
    Yes, the expansion is missing but 400 L x = -100 acc

    How does that tell me what Lamda is?? I don't know.
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    (Original post by maggiehodgson)
    Yes, the expansion is missing but 400 L x = -100 acc

    How does that tell me what Lamda is?? I don't know.
    then x double dot = minus 4 lamda
    so 4 lamda = 100 (you found it at the top of the page as omega squared)
    lamda = 25
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    (Original post by maggiehodgson)
    Yes, the expansion is missing but 400 L x = -100 acc

    How does that tell me what Lamda is?? I don't know.
    In SHM, you have \ddot{x} = -w^2 x, so in this case, you have \ddot{x} = -4 \lambda x

    Hence: 4\lambda  = w^2 \iff \lambda  = \frac{10^2}{4}
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    (Original post by TeeEm)
    then x double dot = minus 4 lamda
    so 4 lamda = 100 (you found it at the top of the page as omega squared)
    lamda = 25
    Oh good grief. I see it now that you've pointed it out.

    Thanks
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    (Original post by Zacken)
    On the penultimate line, you have: 400(0.25g + \lambda x) - 100g= -100\ddot{x}, you should then expand this to get: 400 \lambda x = -100\ddot{x}, you've left it as \lambda x

    Sorry if the LaTeX doesn't work, I'm on my phone and it's very hard to type. :lol:
    Yes , your phone did the trick and so did your response. It makes perfect sense and I'm miffed that I didn't spot it myself.

    Thanks again.
 
 
 
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