Practice question
Question
A stationary wave on a string fixed at both ends has a wavelength of 0.4 m and a frequency of 150 Hz.
What is the wave speed?
Explanation
**Quantization due to boundaries** leads to discrete harmonic series. Frequency difference between harmonics is f₁, so f₃ - f₁ = 2f₁ = v/L. Understanding node-antinode pattern explains resonance and overtones in strings. Speed: v = v λ = 150 × 0.4 = 60 m/s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 60 m/s, illustrating frequency-length-speed interdependence and quantization by boundaries. This aligns with NCERT Class 11 treatment, emphasizing conservation, symmetry and dimensional consistency useful for CBSE, NEET and CUET.
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