Question

Neglecting friction and the effect of small bend near bottom, the time period of oscillations of the particle shown in the figure is $g=10m/s^2$

• A. $\frac{2}{5}[1+\sqrt{\frac{2}{3}}]s$
• B. $\frac{2}{7}[1+\sqrt{\frac{2}{5}}]s$
• C. $\frac{2}{5}[1+\sqrt{\frac{2}{7}}]s$
• D. $\frac{2}{3}[1+\sqrt{\frac{2}{5}}]s$

Answer 1

The correct option is A $\frac{2}{5}[1+\sqrt{\frac{2}{3}}]s$

$\frac{1}{2}g\left(sin{45}^{\circ }\right)t_1^2=\frac{h}{sin{45}^{\circ }}\Rightarrow t_1=\sqrt{\frac{2h}{gsi{n}^2{45}^{\circ }}}=\sqrt{\frac{2}{100\times \frac{1}{2}}}=\frac{1}{5}s\frac{1}{2}g\left(sin{60}^{\circ }\right)t_2^2=\frac{h}{sin{60}^{\circ }}\Rightarrow t_2=\sqrt{\frac{2h}{g}\times \frac{1}{si{n}^2{60}^{\circ }}}=\sqrt{\frac{2}{3}}.\frac{1}{5}s$
Required time period $=2(t_1+t_2)=\frac{2}{5}[1+\sqrt{\frac{2}{3}}]s$