Question

Find the time period of the motion of the particle shown in figure (12-E14). Neglect the small effect of the bend near the bottom.

Answer 1

Let the time taken to travel AB and BC be $t_1$ and $t_2$ respectively.

For part AB, $a_1=gsin{45}^{\circ },$

$s_1=\frac{0.1}{sin{45}^{\circ }}=2m$

Let v = velocity at B

$\therefore v^2-u^2=2a_1{s}_1$

$\Rightarrow v^2=2\times gsin{45}^{\circ }\times \frac{0.1}{sin{45}^{\circ }}=2$

$\Rightarrow v=\sqrt{2}m/s$

$\therefore t_1=\frac{v-u}{a_1}=\frac{\sqrt{2-0}}{\frac{g}{\sqrt{2}}}$

$=\frac{2}{g}=\frac{2}{10}=0.2sec...\left(A\right)$

Again for part BC,

$a_2=gsin{60}^{\circ },$

$u=\sqrt{2}v=0$

$\therefore t_2=\frac{0-\sqrt{2}}{\frac{g}{\left(3\sqrt{2}\right)}}=\frac{2\sqrt{2}}{\sqrt{3}g}$

$=\frac{2\times \left(1.414\right)}{(1.732\times 10)}=0.165sec$

So, time period = $2(t_1+t_2)$

$=2(0.2+0.155)=0.73sec.$