Introduction
We are given two harmonic oscillations:
- x₁ = 8 sin(100πt)
- x₂ = 12 sin(96πt)
These are sinusoidal functions with different frequencies. When superposed, they result in a phenomenon known as beats. The amplitude of the resultant wave varies with time due to interference between the two waves.
Step 1: Beat Frequency
Let the two angular frequencies be:
- ω₁ = 100π rad/s
- ω₂ = 96π rad/s
The beat frequency is the difference of the two frequencies:
fbeat = (ω₁ – ω₂) / 2π = (100π – 96π) / 2π = 4π / 2π = 2 Hz
This means the amplitude varies with a frequency of 2 Hz. Therefore, the time between successive maxima or minima is:
Beat period = 1 / fbeat = 1 / 2 = 0.5 s
Step 2: Resultant Amplitude and Condition for Max/Min
The amplitude of the resultant is given by:
A(t) = √[A₁² + A₂² + 2A₁A₂ cos(Δωt)]
Where:
- A₁ = 8
- A₂ = 12
- Δω = ω₁ – ω₂ = 4π
The amplitude is maximum when cos(Δωt) = 1
⇒ Δωt = 0, 2π, 4π, …
⇒ 4πt = 2πn → t = n/2 seconds
So, maximum amplitude occurs at: t = 0, 0.5, 1.0, 1.5, …
The amplitude is minimum when cos(Δωt) = -1
⇒ Δωt = π, 3π, 5π, …
⇒ 4πt = π(2n + 1) → t = (2n + 1)/4 seconds
So, minimum amplitude occurs at: t = 0.25, 0.75, 1.25, …
Final Answers
- Maximum Amplitude: t = 0, 0.5, 1.0, 1.5, … seconds
- Minimum Amplitude: t = 0.25, 0.75, 1.25, … seconds
Conclusion
When two collinear harmonic oscillations of slightly different frequencies interfere, the resultant wave shows beats. The times at which the amplitude becomes maximum or minimum can be calculated using the beat frequency. This is a key concept in wave interference and applications like tuning musical instruments.