A 3 kg block is connected as shown in the figure. Spring constants of two springs $\mathrm{K}_1$ and…
A 3 kg block is connected as shown in the figure. Spring constants of two springs $\mathrm{K}_1$ and $\mathrm{K}_2$ are $50 \mathrm{Nm}^{-1}$ and $150 \mathrm{Nm}^{-1}$ respectively. The block is released from rest with the springs unstretched, The acceleration of the block in its lowest position is $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$
$10 \mathrm{~ms}^{-2}$
$12 \mathrm{~ms}^{-2}$
$8 \mathrm{~ms}^{-2}$
$8.8 \mathrm{~ms}^{-2}$
Solution
$\mathrm{k}_1=50 \mathrm{Nm}^{-1}, \mathrm{k}_2=150 \mathrm{Nm}^{-1}$
$\mathrm{W}=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{m}}}=\sqrt{\frac{\mathrm{k}_1+\mathrm{k}_2}{\mathrm{~m}}} \sqrt{\frac{50+150}{3}}=\sqrt{\frac{200}{3}} \mathrm{rad} / \mathrm{s}$
Amplitude, $\mathrm{A}=\frac{\mathrm{mg}}{\mathrm{k}_{\mathrm{eq}}}=\frac{3 \times 10}{200}=\frac{3}{20} \mathrm{~m}$
$\therefore$ Acceleration of the block at lowest position,
$A_{\max }=\omega^2 A=\frac{200}{3} \times \frac{3}{20}=10 \mathrm{~m} / \mathrm{s}^2$
