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# M3 Elastic springs (Hooke's Law) watch

1. When an elastic string is stretched to a length of 18cm, the tension in the string is 14N. If the modulus of the string is 28N, find the natural length of the string.
Can anyone please tell me how I can work this out? The extension isn't given so I'm not sure what to do for this.
2. Use T = *x/l (where * is lambda)

But let the extension be [18 - L]
3. thanks
4. Use Hooke's law as normal but instead of using X as the extention use (18-L)
You should then be able to manipulate it and find the natural length, L.
I found it to be 12cm but correct me if i'm wrong!
5. So you get 14 = [28 (18 - L)]/L

Multiply by L: 14L = 28 (18 - L)
Expand : 14L = 504 - 28L
Rearrange: 42L = 504

L = 12cm

Thanks Deepka.
6. (Original post by + polarity -)
So you get 14 = [28 (L - 18)]/L

Multiply by L: 14L = 28 (L - 18)
Expand : 14L = 28L - 504
Rearrange: 14L = 504

L = 36cm
I think the substitution for X is (18-L) rather than (L-18)
7. (Original post by Deepka)
I think the substitution for X is (18-L) rather than (L-18)
Oh, yes!

Schoolboy error.

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Updated: February 7, 2010
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