Question

Find the time period of the motion of the particle shown in figure . Neglect the small effect of the bend near the bottom.

Answer 1

Let t₁ and t₂  be the time taken by the particle to travel distances AB and BC respectively.
Acceleration for part AB, a_1 = g sin 45°
The distance travelled along AB is s_1.

\[\therefore    s_1  = \frac{0 . 1}{\sin  45^\circ} = 2  m\]

Let v be the velocity at point B, and
      u be the initial velocity.

Using the third equation of motion, we have:
v² − u² = 2a₁s₁

\[\Rightarrow  v^2  = 2 \times g  \sin  45^\circ\times \frac{0 . 1}{\sin  45^\circ} = 2\] 

\[ \Rightarrow v = \sqrt{2}  m/s\] 

\[As  v = u +  a_1  t_1 \] 

\[ \therefore  t_1  = \frac{v - u}{a_1}\] 

\[           = \frac{\sqrt{2} - 0}{\frac{g}{\sqrt{2}}}\] 

\[         = \frac{2}{g} = \frac{2}{10} = 0 . 2  \sec    \    (  g = 10   {ms}^{- 2} )\]

For the distance BC,
Acceleration, a₂ =\[-\]gsin 60°

\[\text { Initial  velocity },   u = \sqrt{2}  \] 

\[                                          v   =   0\] 

\[ \therefore   \text { time  period },    t_2  = \frac{0 - \sqrt{2}}{- \frac{g}{\left( 3\sqrt{2} \right)}} = \frac{2\sqrt{2}}{\sqrt{3}g}\] 

\[                                                     = \frac{2 \times \left( 1 . 414 \right)}{\left( 1 . 732 \right) \times 10} = 0 . 163  s\]

Thus, the total time period, t = 2(t₁ + t₂) = 2 (0.2 + 0.163) = 0.73 s