Question

the length of the wire when M1, is hung from it, is l1 and is l2 with both M1 and M2 hanging. the neutral length of wire is

Answer 1

Thank you for asking this question. Here is your answer:

T₁ = m₁g, L₁

T₂ = (m₁ + m₂) g, L₂

L = T₂L₁ - T₁ L₂ / T₂ - T₁

= (m₁ + m₂) g L₁ - m₁ g L₂ / (m₁ + m₂) g - m₁g

= m₁ (L₁ - L₂) - m₂ L₁ / m₂

= m₁/m₂ (L₁ - L₂) + L₁

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Answer 2

Answer: $\frac{M_1}{M_2}(l_1-l_2)+l_1$

Explanation:

Let the Young's modulus of wire be Y

Let the neutral length of wire be l

Let the area of cross section of wire be A

When the mass M₁ is hung, the length is l₁

Then extension in length Δl = l₁-l

$Y=\frac{M_1g/A}{l_1-l/l}$

When the mass M₁+M₂ is hung, the length becomes l₂

$Y=\frac{(M_1+M_2)g/A}{l_2-l/l}$

$\Rightarrow \frac{M_1g/A}{l_1-l/l}=\frac{(M_1+M_2)g/A}{l_2-l/l}$

$\Rightarrow \frac{M_1}{l_1-l}=\frac{M_1+M_2}{l_2-l}\\ \Rightarrow M_1l_2-M_1l=M_1l_1-M_1l+M_2l_1-M_2l\\ \Rightarrow l=\frac{M_1}{M_2}(l_1-l_2)+l_1$

Hence, the neutral length of the wire is $\frac{M_1}{M_2}(l_1-l_2)+l_1$