Question

A body of mass $m$ falls from a height $h$ onto a pan (of negligible mass) of a spring balance as shown. The spring also possess negligible mass and has spring constant $k$. Just after striking the pan, the body starts oscillatory motion in vertical direction of amplitude $A$ and energy $E$. Then

• A. $A=\frac{mg}{k}$
• B. $A=\frac{mg}{k}\sqrt{1+\frac{2kh}{mg}}$
• C. $E=mgh+\frac{1}{2}k{A}^2$
• D. $E=mgh+{\left(\frac{2mg}{2k}\right)}^2$

Answer 1

Answer: (B)

$A=\frac{mg}{k}\sqrt{1+\frac{2kh}{mg}}$

$mg(h+y)=\frac{1}{2}k{y}^2$
$\Rightarrow y=\frac{mg}{k}\pm \frac{mg}{k}\sqrt{1+\frac{2kh}{mg}}$
At equilibrium, $mg=k{y}_0$
$y_0=\frac{mg}{k}$
$\Rightarrow \text{Amplitude A}=y-y_0=\frac{mg}{k}\sqrt{1+\frac{2kh}{mg}}$
Energy of oscillation is
$E=\frac{1}{2}k{A}^2=mgh+\left(\frac{(mg{)}^2}{2k}\right)$