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FM2, Elastics and springs question

An elastic spring has length a1 when supporting a mass m1 and a length a2 when supporting a mass m2. Find the natural length and modulus of elasticity of the spring,

I've tried to rearrange the Hooke's law equation to find the natural length but it cancels out, and I don't even know where to start for the modulus of elasticity. Can anyone help?
Reply 1
Original post by Professor Potato
An elastic spring has length a1 when supporting a mass m1 and a length a2 when supporting a mass m2. Find the natural length and modulus of elasticity of the spring,

I've tried to rearrange the Hooke's law equation to find the natural length but it cancels out, and I don't even know where to start for the modulus of elasticity. Can anyone help?


What did you write down? Hookes law relates the extension and the applied force.
T=λx/l as the equation for Hooke's law. Then I tried doing that when the length is a1, T=m1g, so you get m1g=λ(a1-L)L. I did the same for T when length is a2 but I think this is wrong.
Reply 3
Original post by Professor Potato
T=λx/l as the equation for Hooke's law. Then I tried doing that when the length is a1, T=m1g, so you get m1g=λ(a1-L)L. I did the same for T when length is a2 but I think this is wrong.


So for the first case you have
lambda (a1 - L) / L= m1 g
and similarly for the second case. Simply dividing the two equations would eliminate lambda and you could rearrange for L?
Original post by mqb2766
So for the first case you havelambda (a1 - L) / L= m1 gand similarly for the second case. Simply dividing the two equations would eliminate lambda and you could rearrange for L?

Yeah that works thank you so much. To find λ I'm guessing you just substitute the value of L into T=λx/l?
Reply 5
Original post by Professor Potato
Yeah that works thank you so much. To find λ I'm guessing you just substitute the value of L into T=λx/l?


That should work, just try it? Id be tempted to rearrange it first so something like
lambda = f(L)
then sub for L

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