Introduction
In this problem, we are given the oscillations at two different points on a wave. By analyzing the phase difference and distance between these points, we can determine the wavelength and speed of the wave. This is a typical problem related to wave propagation and phase relationships.
Given
- y₁ = 0.3 sin(4πt) at x = 0
- y₂ = 0.3 sin(4πt + π/8) at x = 1 m
This means the wave at point x = 1 m is ahead in phase by π/8 radians compared to the point at x = 0. The two points are 1 meter apart.
Step 1: Use Phase Difference to Find Wavelength
The general form of a traveling wave is:
y(x, t) = A sin(ωt ± kx)
Here, the phase difference Δϕ between two points a distance Δx apart is:
Δϕ = kΔx = (2π/λ) × Δx
We are told:
- Δϕ = π/8
- Δx = 1 m
Substitute into formula:
π/8 = (2π/λ) × 1
λ = 16 m
Step 2: Calculate Frequency from Angular Frequency
From the wave equation, ω = 4π
f = ω / (2π) = 4π / (2π) = 2 Hz
Step 3: Calculate Speed
Wave speed v is given by:
v = f × λ = 2 × 16 = 32 m/s
Final Answers
- Wavelength (λ): 16 meters
- Speed (v): 32 m/s
Conclusion
Using the phase difference between two known points on a wave, we determined the wavelength and wave speed. This kind of analysis is important for understanding how wave parameters can be derived from observations of oscillatory motion at different spatial points.