JEE Challenger
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PhyMedAdv 2011 • Paper 2

SHM of Charged Block in Uniform Electric Field

A wooden block performs SHM on a frictionless surface with frequency, ν0\nu_0. The block carries a charge +Q+Q on its surface. If now a uniform electric field E⃗\vec{E} is switched-on as shown, then the SHM of the block will be

Diagram 0

Options

A

of the same frequency and with shifted mean position.

Correct
B

of the same frequency and with the same mean position.

C

of changed frequency and with shifted mean position.

D

of changed frequency and with the same mean position.

Topics & Concepts

OscillationsSHMElectric Field

Step-by-Step Solution

To analyze the Simple Harmonic Motion (SHM) of the wooden block, let us consider the forces acting on it before and after the electric field is applied.

1. Before the Electric Field is Applied:

Let mm be the mass of the wooden block and kk be the spring constant of the spring. The equation of motion for the block executing SHM about its natural length (mean position x=0x = 0) is:
md2xdt2=−kxm \frac{d^2x}{dt^2} = -kx

The angular frequency ω0\omega_0 and the natural frequency ν0\nu_0 of the oscillation are given by:
ω0=km\omega_0 = \sqrt{\frac{k}{m}}
ν0=12πkm\nu_0 = \frac{1}{2\pi}\sqrt{\frac{k}{m}}

2. After the Electric Field is Applied:

When a uniform electric field E⃗\vec{E} directed horizontally to the right is switched on, an additional constant electrostatic force acts on the charge +Q+Q:
F⃗e=QE⃗\vec{F}_e = Q\vec{E}

The net force acting on the block at a displacement xx from the uncompressed state of the spring becomes:
Fnet=−kx+QEF_{\text{net}} = -kx + QE

Thus, the differential equation of motion is:
md2xdt2=−kx+QEm \frac{d^2x}{dt^2} = -kx + QE

Rearranging the terms:
md2xdt2=−k(x−QEk)m \frac{d^2x}{dt^2} = -k \left(x - \frac{QE}{k}\right)

3. Shift in Mean Position:

At the new equilibrium (mean) position, the net force on the block is zero:
−kxmean+QE=0  ⟹  xmean=QEk-kx_{\text{mean}} + QE = 0 \implies x_{\text{mean}} = \frac{QE}{k}

Thus, the mean position shifts to the right by a distance of Δx=QEk\Delta x = \frac{QE}{k}.

4. Frequency of Oscillation:

Let x′=x−QEkx' = x - \frac{QE}{k} represent the displacement from the new mean position. Differentiating twice with respect to time gives:
d2x′dt2=d2xdt2\frac{d^2x'}{dt^2} = \frac{d^2x}{dt^2}

Substituting this back into the equation of motion yields:
md2x′dt2=−kx′m \frac{d^2x'}{dt^2} = -kx'

This is the standard equation of Simple Harmonic Motion. The new angular frequency ω′\omega' and frequency ν′\nu' are:
ω′=km=ω0\omega' = \sqrt{\frac{k}{m}} = \omega_0
ν′=12πkm=ν0\nu' = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \nu_0

Since a constant force does not alter the effective force constant kk or mass mm, the frequency of oscillation remains unchanged, while the mean position is shifted.

Conclusion:

The SHM of the block will be of the same frequency and with shifted mean position.

Correct Option: A

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