Question

Find the time period of the motion of the particle shown in figure. Neglect the effect of the bend at the bottom. Use $g=10m{s}^{-2}.$

• A. 0.15 s
• B. 0.97 s
• C. 0.25 s
• D. 1.23 s

Answer 1

Answer: (B)

0.97 s

After covering $0.1$ height or $\frac{0.1}{\sin 45}$ along the incline , the velocity of particle is

$v^2=2g\sin 45\frac{0.1}{\sin 45}=2$

Time taken be $t$ secs (say)

$v=g\sin 45t$

$t=0.2s$

Now the particle moves along a slope with $30$ degree incline . Time taken before it stops and reverse the direction be $t_1$

$v=g\sin 30t_1$

$t_1=0.28s$

Now the particle will again come back and reach point A.

Total time elapsed after starting from point A and coming back to point A

i.e. the time period of motion $=2(t+t_1)=0.97s$