Question

A block of mass m containing a net positive charge q is placed on a smooth horizontal table which terminates in a vertical wall as shown in figure . the distance of the block from the wall is d .A horizontal electric field E toward right it switched on . assuming electric collisions (if any ) find the time period of the resulting oscillatory motion. Is it a simple harmonic motion ​

Answer 1

Answer ❤

Acceleration of block is a = m

qE

Time taken to reach the wall is equal to time taken from wall to reach the starting position since the collision is elastic.

d = 2

1

m

qE

t

2

= t = qE

2 dm

Hence, time taken to reach initial point is T= 2t

Time of oscillation = T= qE

8dm

Hope its help u ❤

Answer 2

$\bigstar\huge\boxed{\mathtt{Solution:}}$

The block undergoes oscillatory motion and not simple harmonic motion

Reason :

we know that ,

$\implies\mathtt{a= -{\omega}^{2} x}$

From this we can conclude that acceleration and displacement are directly proportional to each other but are in opposite direction to each other

When the block is moving towards the wall the displacement and the acceleration both are acting in the same direction but when the block is moves away from the wall the displacement is and the acceleration both are in opposite directions

Hence , the movement of the block is oscillatory and not simple harmonic

As the acceleration is constant  ,we can use the second kinematic equation in order to solve this question,

$\implies\mathtt{s=ut+\dfrac{1}{2} at^{2} }$

• Initial velocity of the block = 0 m/s

• acceleration of the block = q E/m

• Distance covered = d

$\implies\mathtt{d=\dfrac{1}{2} \dfrac{qE}{m} t^{2} }$

$\implies\mathtt{ t^{2}= \dfrac{2md}{qE} }$

$\implies\mathtt{ t= \sqrt{\dfrac{2md}{qE} }}$

As the collision is completely elastic the time taken to go to the wall and come back (original position) will be the same

Hence the total time taken ,

$\implies\mathtt{T= 2t }$

$\implies\mathtt{T= 2\sqrt{\dfrac{2md}{qE} } }$