Question

A solid ball of mass m is allowed to fall from a height h to a pan suspended with a spring of spring constant k. Assume the ball does not rebound and pan is massless,then amplitude of the oscillation is

• A. $\frac{mg}{k}$
• B. $\frac{mg}{k}+{\left(\frac{2hk}{mg}\right)}^{1/2}$
• C. $mg\sqrt{1+\frac{1+2hk}{mg}}$
• D. $\frac{mg}{k}[1+\sqrt{1+\frac{2hk}{mg}}]$

Answer 1

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

Here the gravitational potential energy of the ball converts into the potential energy of the spring. So we have
$mg(h+x)=\frac{1}{2}k{x}^2$
$\Rightarrow x^2-\left(\frac{2mg}{k}\right)x-\frac{2mgh}{k}=0$
after solving the above quadratic equation, We have
$x=\frac{mg}{k}+\frac{mg}{k}\sqrt{1+\frac{2hk}{mg}}$
Note: $x=\frac{mg}{k}-\frac{mg}{k}\sqrt{1+\frac{2hk}{mg}}$ is negative and cant be the solution because extension can not be negative. and why it is negative, it is explained below
$1+\frac{2hk}{mg}>1$ Since $h$,$k$,$m$ and $g$ are positive
$\Rightarrow \sqrt{1+\frac{2hk}{mg}}>1\Rightarrow \frac{mg}{k}\left(\sqrt{1+\frac{2hk}{mg}}\right)>\frac{mg}{k}\Rightarrow \frac{mg}{k}-\frac{mg}{k}\left(\sqrt{1+\frac{2hk}{mg}}\right)<0$