Question

An elastic cord of constant K and length L is hung from point A having a massless lock at the other end. A smooth ring of mass M falls from point A, the maximum elongation of cord is

• A. $\frac{Mg}{K}\left( 1 + \frac{1 + 2KL}{Mg} \right)^{1/2}$
• B. $\frac{Mg}{K}\left( 1 - \left( 1 - \frac{2KL}{Mg} \right)^{1/2} \right)$
• C. $\frac{MgL}{K}$
• D. $\frac{Mg}{K}\left( 1 + \left( 1 + \frac{2KL}{Mg} \right)^{1/2} \right)$

Answer 1

Answer: (D)

$\frac{Mg}{K}\left( 1 + \left( 1 + \frac{2KL}{Mg} \right)^{1/2} \right)$

Answer 2

Let the extension in cord be KL. Using conservation of energy, we get

Mg(L + δL) = $\frac{1}{2}$K(δL)²

K(δL) + 2 – 2 MgL – 2MgδL = 0

or K(δL)² – 2 MgδL – 2MgL = 0

or δL = $\frac{2Mg \pm \sqrt{4M^{2}g^{2} + 4K \times 2MgL}}{2K}$ (By solving the quadratic equation)

i.e., δL = $\frac{Mg}{K}\left( 1 + \sqrt{1 + \frac{2KL}{Mg}} \right)$