Superposition of Waves (Interference & Beats)
AP EAMCET – Physics (Class‑12)
Topic: Superposition of Waves: Interference, Beats
1. Simple Explanation
-
Superposition Principle:
When two (or more) waves occupy the same region of space, the resultant displacement at any point and time is the sum of the individual displacements. -
Mathematically:
[ y(x,t)=y_1(x,t)+y_2(x,t)+\dots ] -
Consequences:
- Constructive interference → amplitudes add → larger amplitude.
- Destructive interference → amplitudes cancel → smaller or zero amplitude.
- Standing waves → superposition of two waves travelling in opposite directions.
-
Beats:
When two waves have almost the same frequency, their superposition produces a modulation in amplitude—beats—with frequency equal to the difference of the two frequencies.
2. Key Points
| Aspect | Details |
|---|---|
| Superposition | Linear addition of displacements: valid for linear media (small amplitude). |
| Constructive Interference | Phase difference Δφ = 2πn (n = 0,1,2…) → maximum amplitude (A_{max}=A_1+A_2). |
| Destructive Interference | Δφ = (2n+1)π → minimum amplitude (A_{min}= |
| Standing Wave | (y(x,t)=2A\sin(kx)\cos(\omega t)) – nodes at sin(kx)=0, antinodes at cos(ωt)=±1. |
| Beat Frequency | (f_{\text{beat}}= |
| Beat Amplitude | (A_{\text{beat}}=2A\cos\left(\frac{Δ\phi}{2}\right)). |
| Wave Equations | (y_1=A\sin(kx-\omega t)), (y_2=A\sin(kx-\omega t+Δφ)). |
| Applications | Musical tuning, radar, interference patterns, radio broadcasting. |
3. Worked Example
Problem
Two transverse waves of the same amplitude (A=2,\text{cm}) travel along the same string:
[ y_1=A\sin(kx-\omega t),\qquad y_2=A\sin(kx-\omega t+\tfrac{\pi}{3}) ]
- Find the resultant wave equation.
- Determine the amplitude at a point where (k x = \frac{\pi}{2}).
- If the waves have frequencies (f_1=100,\text{Hz}) and (f_2=108,\text{Hz}), calculate the beat frequency.
Solution
- Superposition
[ \begin{aligned} y &= y_1 + y_2\ &= A\sin(kx-\omega t)+A\sin(kx-\omega t+\tfrac{\pi}{3})\ &= 2A\cos!\left(\frac{\tfrac{\pi}{3}}{2}\right)\sin!\left(kx-\omega t+\frac{\tfrac{\pi}{3}}{2}\right)\ &= 2A\cos!\left(\frac{\pi}{6}\right)\sin!\left(kx-\omega t+\frac{\pi}{6}\right) \end{aligned} ]
Since (\cos(\pi/6)=\sqrt{3}/2),
[ \boxed{y = 2A\left(\frac{\sqrt{3}}{2}\right)\sin!\left(kx-\omega t+\frac{\pi}{6}\right)=A\sqrt{3},\sin!\left(kx-\omega t+\frac{\pi}{6}\right)} ]
So the resultant amplitude is (A_{\text{res}} = A\sqrt{3}).
- Amplitude at (k x = \pi/2)
The amplitude is independent of (x) and (t) for this case (since we already expressed the resultant as a single sine). Thus,
[ A_{\text{res}} = A\sqrt{3} = 2,\text{cm}\times1.732 = 3.464,\text{cm} ]
- Beat Frequency
[ f_{\text{beat}} = |f_1-f_2| = |100-108| = 8,\text{Hz} ]
So the combined wave will have an envelope oscillating at 8 Hz.
4. Quick Revision Summary
| Concept | Formula / Condition | Quick Check |
|---|---|---|
| Superposition | (y = y_1 + y_2) | Add the equations. |
| Constructive | Δφ = 2πn | (A_{\text{res}} = A_1 + A_2) |
| Destructive | Δφ = (2n+1)π | (A_{\text{res}} = |