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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.
Scan0010.pdf360.5KB
Reply 1
Avatar for TeeEm
TeeEm
19
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
(edited 8 years ago)
Reply 2
Avatar for Zacken
Zacken
22
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+λx)−100g=−100x¨400(0.25g + \lambda x) - 100g= -100\ddot{x}, you should then expand this to get: 400λx=−100x¨400 \lambda x = -100\ddot{x}, you've left it as λx\lambda x

Sorry if the LaTeX doesn't work, I'm on my phone and it's very hard to type. :lol:
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.
Reply 4
Avatar for TeeEm
TeeEm
19
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
Reply 5
Avatar for Zacken
Zacken
22
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 x¨=−w2x\ddot{x} = -w^2 x, so in this case, you have x¨=−4λx\ddot{x} = -4 \lambda x

Hence: 4λ=w2  ⟺  λ=10244\lambda = w^2 \iff \lambda = \frac{10^2}{4}
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
Original post by Zacken
On the penultimate line, you have: 400(0.25g+λx)−100g=−100x¨400(0.25g + \lambda x) - 100g= -100\ddot{x}, you should then expand this to get: 400λx=−100x¨400 \lambda x = -100\ddot{x}, you've left it as λx\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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