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# A body of mass $M$ and charge $q$ is connected to a spring of spring constant $k$. It is oscillating along x-direction about its equilibrium position, taken to be at $x = 0$, with an amplitude $A$. An electric field $E$ is applied along the x-direction. Which of the following statements is correct?(A) The total energy of the system is $\dfrac{1}{2}m{\omega ^2}{A^2} +\dfrac{1}{2}\dfrac{{{q^2}{E^2}}}{k}$(B) The new equilibrium position is at a distance: $\dfrac{{2qE}}{k}$ from $x = 0$(C) The new equilibrium position is at a distance $\dfrac{{qE}}{{2k}}$ from $x = 0$(D) The total energy of the system is $\dfrac{1}{2}m{\omega ^2}{A^2} -\dfrac{1}{2}\dfrac{{{q^2}{E^2}}}{k}$

Last updated date: 11th Sep 2024
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Hint:Initially the body of mass $M$ is oscillating about $x = 0$, this oscillation is simple harmonic oscillation because the body is attached to spring. Now, an electric field is applied in the x-direction.

This electric field will apply force on the body as it is charged. Consider the force on the body, find the acceleration of the body and look at changes occurring as an electric field is applied, accordingly determine the new equilibrium position and find the energy of the system.

Complete step by step solution:
The electric field applied is constant.
Consider the body to be moving in the positive x-direction and let the distance of the body from the mean position $x = 0$ be $x$.

The forces on the body will be due to the spring towards the mean position (negative direction) and due to the electric field in positive x-direction.

$F = - kx + qE$, because the force on a charged particle due to an electric field is given as charge multiplied by the electric field.
This force has to be equal to mass times acceleration according to laws of motion, we have,
$Ma = - kx + qE \\ a = - \dfrac{k}{M}x + \dfrac{{qE}}{M} \\ \therefore a = - \dfrac{k}{M}\left( {x - \dfrac{{qE}}{k}} \right) \\$

As you can see that this equation resembles the equation of SHM ($a = - \left( {\dfrac{k}{M}} \right)x$). Hence, it is still an SHM but with a different mean position. The mean position is where the acceleration of the body becomes zero. So, we have $a = - \dfrac{k}{M}\left( {x - \dfrac{{qE}}{k}} \right) = 0 \to x = \dfrac{{qE}}{k}$. Therefore, the new mean/equilibrium position is at a distance of $x = \dfrac{{qE}}{k}$ from $x = 0$.

Now, let us find the total energy of the system.
Consider the particle at its equilibrium position, the acceleration will be zero, the velocity will be maximum and is equal to $v = \omega A$ where $\omega$ is the angular frequency and $A$ is equal to the amplitude. Also, we have a stretch of $\dfrac{{qE}}{k}$ in the spring (since, the position where the stretch is zero is $x = 0$). Therefore, the total energy is given by $T.E = K.E + {E_{spring}}$
$T.E = \dfrac{1}{2}m{v^2} + \dfrac{1}{2}k{x^2} \\ T.E = \dfrac{1}{2}m{\left( {\omega A} \right)^2} + \dfrac{1}{2}k{\left( {\dfrac{{qE}}{k}} \right)^2} \\ T.E = \dfrac{1}{2}m{\omega ^2}{A^2} + \dfrac{1}{2}\dfrac{{{q^2}{E^2}}}{k} \\$

Here, for our ease, we have considered the total energy at the new equilibrium position. You can consider any position you want, you will get the same answer.

Therefore, the total energy of the system is $\dfrac{1}{2}m{\omega ^2}{A^2} + \dfrac{1}{2}\dfrac{{{q^2}{E^2}}}{k}$ and the new equilibrium position is at a distance
$\dfrac{{qE}}{k}$ from $x = 0$.

Option (A) is correct.

Note:Whenever a constant electric field is applied, it only changes the mean position of the oscillating charged particle. It does not affect anything else. Remember how we used the force on the body to determine the new equilibrium position. Also keep in mind the conditions at mean position, like the values of acceleration and velocity. The total energy of the system is given as the sum of kinetic energy and the potential energy. In this case, the potential energy is the energy stored in the spring.