Question

A mass $m$ fall on spring constant $k$ and negligible mass from a height $h$. Assuming it sticks to the pan and executes simple harmonic motion, the maximum height upto which the pan will rise is:

• A. $\frac{mg}{k}$
• B. $\frac{mg}{k}[\sqrt{1+\frac{2kh}{mg}}-1]$
• C. $\frac{mg}{k}[\sqrt{1+\frac{2kh}{mg}}+1]$
• D. $\frac{mg}{k}[\sqrt{1+\frac{kh}{mg}}-1]$

Answer 1

The correct option is C $\frac{mg}{k}[\sqrt{1+\frac{2kh}{mg}}+1]$

When the mass m falls on the spring from a height, let the length of the spring compressed be x.

Total distance by which the mass m falls is: $(h+x)$
Loss of PE of the mass is equal to the energy stored in the spring.
$\therefore mg(h+x)=\frac{1}{2}k{x}^2$
$\therefore x^2-\left(\frac{2mg}{k}\right)x-\frac{2mgh}{k}=0$
$\therefore x=\frac{\frac{2mg}{k}\pm \sqrt{(\frac{2mg}{k}{)}^2+\frac{8mgh}{k})}}{2}$
$\therefore x=\frac{mg}{k}\pm \sqrt{(\frac{mg}{k}{)}^2+\frac{2mgh}{k})}=\frac{mg}{k}(1+\sqrt{(1+\frac{2kh}{mg})})$
We ignore the -ve sign, since it will give a negative x which is not a physical situation.
Thus, the maximum height to which the mass will rise is: $\frac{mg}{k}[1+\sqrt{(1+\frac{2kh}{mg})}]$