Q.A charged particle oscillates about its mean equilibrium position with a frequency of $10^{9}\ \text{Hz}$. What is the frequency of the electromagnetic waves produced by the oscillator?

The key idea is that an oscillating charge produces electromagnetic waves of the same frequency as its own oscillation.

• The charge oscillates at $f = 10^{9}\ \text{Hz}$.
• The electromagnetic waves generated by the accelerating charge have the same frequency as the source oscillation. …

The key idea here is beautifully simple: an oscillating electric charge is the source of electromagnetic waves, and the wave it produces cannot have a frequency different from the source's own motion. Let's see why.

When a charged particle (like an electron) moves back and forth, it creates a changing electric field. A changing electric field, by Maxwell's laws, generates a magnetic field, and a changing magnetic field regenerates an electric field — and so on. This self-sustaining chain propagates outward as an electromagnetic wave.

The crucial point: the source's oscillation drives the wave. The electric field at any point near the charge varies exactly as the charge's position varies. If the charge completes one full back-and-forth cycle in time $T$, the electric field at a fixed point also completes one full cycle in the same $T$. Therefore, the wave's frequency $f$ must equal the source's frequency $f_0$.

1. Identify the source frequency. The problem states the charged particle oscillates with frequency $f_0 = 10^{9}\ \text{Hz}$.

2. Relate source motion to wave production. The oscillating charge acts as an antenna. Its acceleration (since oscillation involves acceleration) produces electromagnetic radiation. The radiated field's time variation is locked to the source's motion — there is no mechanism for the wave to "wiggle" faster or slower than the charge itself.

3. Apply the fundamental relation. For any electromagnetic wave produced by an oscillating dipole (which is what a single oscillating charge approximates), the wave frequency equals the oscillation frequency of the dipole. So: …

Method: Direct Application of the Source-Frequency Principle

Concept: An oscillating charged particle produces electromagnetic waves of the same frequency as its own oscillation. This is a fundamental result from Maxwell's equations — the source frequency equals the radiated wave frequency.

Steps

1. Identify the source frequency The charged particle oscillates at:

2. Apply the principle An accelerating charge radiates electromagnetic waves. The frequency of the radiated wave is exactly equal to the frequency of oscillation of the source charge. $\boxed{f_{\text{wave}} = f_{\text{source}}}$ …

Let’s break this down — it’s a short but powerful concept question from the Electromagnetic Waves chapter in Class 12 Physics.

🔍 The Core Concept

When a charged particle oscillates, it produces electromagnetic waves of the same frequency as the oscillation.

So here:

• Oscillation frequency of charged particle = $10^{9}\ \text{Hz}$
• Frequency of EM waves produced = $10^{9}\ \text{Hz}$

That’s the correct answer.

✗ Common Mistakes Students Make

1. Thinking the frequency changes due to speed of light

• Mistake: Students think: “EM waves travel at $3 \times 10^8$ m/s, so frequency must change.”
• Why it’s wrong: Frequency is determined by the source (the oscillator). Speed changes in different media, but frequency remains constant.
• How to avoid: Remember — frequency is source-dependent, not medium-dependent.

2. Confusing frequency with wavelength

• Mistake: Trying to use $c = f \lambda$ and incorrectly calculating a new frequency.
• Why it’s wrong: $c = f \lambda$ relates speed, frequency, and wavelength — but frequency $f$ is still the oscillator’s frequency.
• How to avoid: Use $c = f \lambda$ only to find wavelength, not to change frequency.

3. Assuming the particle’s amplitude affects frequency

• Mistake: Thinking larger oscillation amplitude means higher EM wave frequency.
• Why it’s wrong: Amplitude affects intensity (energy), not frequency. …