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The question is related to the situation:

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.

Now one has to find the total energy of the system after the electric field is applied. Wouldn't different initial conditions lead to different outcome?

For example, when electric field is applied ,two different initial conditions could be :

I1) the mass is passing through mean/ equilibrium point
I2) the mass is at the extreme end of the ongoing S.H.M.

When Electric field is switched on , there is no material contact with the mass attached to the spring. But the answers I get is different for the cases mentioned above. Is there any field interaction and momentum transfer concept involved here?

Attempt at solving:

The force on mass m could be written as:

$F=-kx+qE$

$m a=-kx+qE$

$a=-kx/m+qE/m$

$\frac {dv}{dt}=-\frac{kx}{m}+\frac{qE}{m}$

Using $\frac {dv}{dt}= v\frac{dv}{dx}$ and rearranging yields:

$vdv=-\frac{kx}{m} dx+\frac{qE}{m} dx $

Integrating both sides yields:

$\frac{v^2}{2}=-\frac{kx^2}{2m}+\frac{qEx}{m}+C -- --- (1)$

C= constant of integration

Equation (1) above is true for any initial condition. Value of C is determined by using a given initial condition.

I1 / Initial condition 1: the mass is passing through mean point , i.e. x=0, this implies

$v_0=ωA ; where ω=\sqrt{\frac{k}{m}}$

Substituting $v_0=ωA$ in Eq (1) yields the value of C as

$\frac{(ωA)^2}{2}=C$

Thus:

$\frac{v^2}{2}=-\frac{kx^2}{2m}+\frac{qEx}{m}+\frac{ω^2 A^2}{2}$

Using $ω^2=k/m$

$\frac{v^2}{2}=-\frac{kx^2}{2m}+\frac{qEx}{m}+ \frac{kA^2}{2m} ----(2) $

Rearranging terms,

$v^2=-\frac{k}{m} (x-\frac{qE}{k})^2+ \frac{k}{m} [(\frac{qE}{k})^2+A^2 ] ---- (3)$

I2/Initial condition 2: the mass is at right extreme point , i.e. x=A, this implies

$v_A=0$

Substituting $v_A=0$ in (1) yields the value of C.

$\frac{0^2}{2}=-\frac{kA^2}{2m}+\frac{qEA}{m}+C$

$C=\frac{kA^2}{2m}-\frac{qEA}{m}$

Thus:

$\frac{v^2}{2}=-\frac{kx^2}{2m}+\frac{qEx}{m}+\frac{kA^2}{2m}-\frac{qEA}{m} -----(4)$

rearranging equation (4) yields:

$v^2= -\frac{k}{m}[(x-\frac{qE}{k})^2-(A-\frac{qE}{k})^2] ----- (5)$

Expression for v differs in equation (3) and equation (5), clearly total energy differs for differing initial conditions.

SHM frequency remains the same as evident.

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  • $\begingroup$ Adding a constant to the potential doesn't change the outcome. $\endgroup$
    – Vincent Thacker
    Commented 2 days ago
  • $\begingroup$ This problem is analogous to having a mass oscillating at the end of a spring without a gravitational force acting and comparing that situation if a gravitational force is acting. $\endgroup$
    – Farcher
    Commented 2 days ago
  • $\begingroup$ Are you asking if the answer changes if the electric field is turned on when the mass is at maximum x vs. minimum x? $\endgroup$
    – mmesser314
    Commented 2 days ago
  • $\begingroup$ yes, exactly @mmesser314 $\endgroup$
    – user171297
    Commented 2 days ago
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    $\begingroup$ You are right. But it does not look like the question was trying to ask about initial conditions. I think the author missed that point. The question is trying to get you to show that if the amplitude is A, the energy is the same whether or not the electric field is present. $\endgroup$
    – mmesser314
    Commented 2 days ago

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