Question

Find the time period of the motion of the particles shown in figure. 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 , $a_1=gsin{45}^0$ $s_1=\frac{0.1}{sin{45}^o}=2m$

Let $v = velocity at B

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

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

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

$\therefore$ $t_1=\frac{v-u}{a_1}=\frac{\sqrt{2}-0}{\frac{g}{\sqrt{2}}}=\frac{2}{g}=0.2sec$

Again for part BC, $a_2=-gsin{60}^o$, $u=\sqrt{2}m/s$

$\therefore$ $t_1=\frac{v-u}{a_1}=\frac{0-\sqrt{2}}{\frac{-g\sqrt{3}}{2}}=\frac{2\sqrt{2}}{\sqrt{3}g}=0.165sec$

So, the time period = $2(t_1+t_2)=2(0.2+0.165)=0.71sec$