Physics - SHM Question with Solution | TestHub

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$=x_0-x_1$

$=\frac{mg}{k} \sqrt{\left(1+\frac{2hk}{mg}\right)}$