Find the Time Period of the Motion of the Particle Shown in Figure . Neglect the Small Effect of the Bend Near the Bottom. - Physics

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Sum

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

Solution

Let t1 and t2  be the time taken by the particle to travel distances AB and BC respectively.
Acceleration for part AB, a1 = sin 45°
The distance travelled along AB is s1.

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

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

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Chapter 12: Simple Harmonics Motion - Exercise [Page 254]

APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 12 Simple Harmonics Motion
Exercise | Q 29 | Page 254

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