Question

A load of mass $m$ falls from a height $h$ on to the scale pan hung from a spring, as shown in the figure. If the spring constant is $k$ , mass of the scale pan is zero and the mass $m$ does not bounce, relative to the pan, then the amplitude of vibration is

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

Answer 1

Answer: (B)

According to conservation principle,
$mgh=\frac{1}{2}k{x}_0^2-mg{x}_0$
where $x_0$ is maximum elongation in spring (when particle is in its lowest extreme position)
or $\frac{1}{2}k{x}_0^2-mg{x}_0-mgh=0$
or $x_0^2-\frac{2mg}{k}{x}_0-\frac{2mg}{k}h=0$
$\therefore$ $x_0=\frac{\frac{2mg}{k}\pm \sqrt{\left[{\left(\frac{2mg}{k}\right)}^2+4\times \frac{2mg}{k}h\right]}}{2}$
$\because$ Amplitude = elongation in spring for lowest extreme position - elongation in spring for equilibrium position $=x_0-x_1$
$=\frac{mg}{k}\sqrt{\left(1+\frac{2hk}{mg}\right)}$