Question

A particle of mass $ m $ and charge $ Q $ is placed in an electric field $ E $ which varies with time t as $ E = {E_0}\sin \omega t $ . It will undergo simple harmonic motion of amplitude:
A. $ \dfrac{{Q{E_0}^2}}{{m{\omega ^2}}} $
B. $ \dfrac{{Q{E_0}}}{{m{\omega ^2}}} $
C. $ \sqrt {\dfrac{{Q{E_0}}}{{m{\omega ^2}}}} $
D. $ \dfrac{{Q{E_0}}}{{m\omega }} $

Answer 1

Hint: Use the standard formula for finding the force applied on a charged particle due to an electric field. To find the amplitude, we need to integrate the expression. It requires another integration because we want to obtain the amplitude of the motion.

Complete step by step solution:
We have,
  $ E = {E_0}\sin \omega t $
Now we need to calculate the force on the given charged particle of mass $ m $ and charge $ Q $ .
  $
  F = m \times a = QE = Q{E_0}\sin \omega t \\
  m\dfrac{{dv}}{{dt}} = Q{E_0}\sin \omega t \\
  $
After integrating both sides with respect to $ dt $ , we get
  $
  m\int {dt} = Q{E_0}\int {\sin \omega t} .dt \\
  mv = \dfrac{{ - Q{E_0}}}{\omega }\cos \omega t + {C_1} \\
  $
Thus, from this,
  $ v = \dfrac{{ - Q{E_0}}}{{m\omega }}\cos \omega t + {C_1} = \dfrac{{dx}}{{dt}} $
After integrating both sides again, we get
  $
  \int {dx} = \dfrac{{ - Q{E_0}}}{{m\omega }}\int {\cos \omega t} .dt + \int {{C_1}.dt} \\
  x = \dfrac{{Q{E_0}}}{{m{\omega ^2}}}\sin \omega t + {C_1}t + {C_2} \\
  $
Here, the first term on the right provides the SHM part, and the coefficient of the first term is the amplitude.
Hence, option B is the correct answer.

Note:
In mechanics and physics, simple harmonic motion is a special category of periodic motion, in which the restoring force on the moving object is directly proportional to the object's displacement magnitude and acts towards the object's equilibrium.