Question

A block of charge q and mass m is connected to a spring of constant k. An electric field E exists parallel to the ground. The block is released from rest from a unstreched spring. Find Maximum displacement?

Answer 1

`(2Eq)/m`

Explanation:

Newton's 2nd law:

• `F = ma`

Here:

• `F = Eq - kx`

The spring linearly opposes displacement from the equilibrium position, hence the negative term and the harmonic oscillation.

Hence, equation of motion:

`Eq - kx = m ddot x`

• `implies ddot x + k/mx = (Eq)/m`

General solution:

• `x = A cos omega t + B sin omega t + (Eq)/k`, where `qquad omega^2 = k/m`

With IV's:

• `x(0) = 0`

`implies x = B sin omega t + (Eq)/k (1- cos omega t)`

• `x'(0) = 0`

`x' = omega B cos omega t + omega (Eq)/k sin omega t implies B = 0`

So the governing equation is:

• `x = (Eq)/k (1- cos sqrt(k/m) t)`

Because `-1 le cos theta le 1`:

• `0 lt x lt (2Eq)/k`