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

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#### Solution

Let *t*_{1} and *t*_{2} 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:*v ^{2}* −

*u*

^{2}= 2

*a*

_{1}

*s*

_{1}

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

\[ \Rightarrow v = \sqrt{2} m/s\]

\[As v = u + a_1 t_1 \]

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

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

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

For the distance BC,

Acceleration, *a*_{2} =\[-\]*g*sin 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(*t*_{1} + *t*_{2}) = 2 (0.2 + 0.163) = 0.73 s

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