Practice question
Question
A wave on a string has an amplitude of 0.015 m and a frequency of 40 Hz. What is the maximum transverse
speed of a particle?
Explanation
**Interference of waves** produces enhancement or cancellation based on phase. Two equal amplitude waves out of phase by π cancel completely, A = 0, while in-phase superposition doubles amplitude to 2a, demonstrating energy redistribution without violation of conservation. ω = 2π v = 2π × 40 = 80π rad/s . Max speed: vₘₐₓ = a ω = 0.015 × 80π ≈ 3.77 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 3.8 m/s, illustrating frequency-length-speed interdependence and quantization by boundaries.
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