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Question: A particle of mass \( m \) and charge \( Q \) is placed in an electric field \( E \) which varies wi...

A particle of mass mm and charge QQ is placed in an electric field EE which varies with time t as E=E0sinωtE = {E_0}\sin \omega t . It will undergo simple harmonic motion of amplitude:
A. QE02mω2\dfrac{{Q{E_0}^2}}{{m{\omega ^2}}}
B. QE0mω2\dfrac{{Q{E_0}}}{{m{\omega ^2}}}
C. QE0mω2\sqrt {\dfrac{{Q{E_0}}}{{m{\omega ^2}}}}
D. QE0mω\dfrac{{Q{E_0}}}{{m\omega }}

Explanation

Solution

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=E0sinωtE = {E_0}\sin \omega t
Now we need to calculate the force on the given charged particle of mass mm and charge QQ .
F=m×a=QE=QE0sinωt mdvdt=QE0sinωt  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 dtdt , we get
mdt=QE0sinωt.dt mv=QE0ωcosωt+C1  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=QE0mωcosωt+C1=dxdtv = \dfrac{{ - Q{E_0}}}{{m\omega }}\cos \omega t + {C_1} = \dfrac{{dx}}{{dt}}
After integrating both sides again, we get
dx=QE0mωcosωt.dt+C1.dt x=QE0mω2sinωt+C1t+C2  \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.