Question

A block of mass $m$ having charge $+Q$ is released from rest on a smooth surface at the end of which is an elastic vertical wall (see figure). The time period of motion of the block will be [$E$ is uniform electric field]

• A. (A) $T = \sqrt{\frac{2 d m}{Q E}}$
• B. (B) $T = \sqrt{\frac{8 d m}{Q E}}$
• C. (C) $T = 2 \sqrt{\frac{2 d m}{Q E}}$
• D. (D) $T = \sqrt{\frac{d m}{Q E}}$

Answer 1

Answer: (B), (C)

$\sqrt{\frac{8dm}{QE}}$

Answer 2

The block experiences a constant force $QE$ towards the wall, resulting in constant acceleration $a = QE/m$. The time taken to reach the wall from rest, covering distance $d$, is $t_1 = \sqrt{2d/a}$. Due to the elastic collision, the block rebounds with the same speed and decelerates back to its starting position, taking an equal time $t_2 = \sqrt{2d/a}$. The total time period of oscillation is $T = t_1 + t_2 = 2\sqrt{2d/a} = 2\sqrt{2dm/(QE)} = \sqrt{8dm/(QE)}$.