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.
Figure

Answer 1

The time period of the motion of the particle is $T=0.71 s.$

Explanation:

• The time period of the motion of the particle can be written as $2\left(t_{1}+t_{2}\right)$ where $t_1$ is the time taken for the particle to travel from A to B and $t_{2}$ is the time taken to travel from B to A.  
• We know that, $a_{1}=g \sin 45^{\circ}$ and $a_{2}=-g \sin 60^{\circ}$. Also, we can write $t_{1}=\frac{v-u}{a_{1}}=\frac{2}{10}=0.2 \mathrm{s}$ and  $t_{2}=\frac{v-u}{a_{2}}=\frac{2 \sqrt{2}}{\sqrt{3 g}}=0.165 \mathrm{s}$ where v is the velocity at B.  
• Thus, on substituting the known values, we get, $2\left(t_{1}+t_{2}\right)=T=0.71 s$.