Tuning a String: Which Variable to Change
A guitar string of length 0.65 m and linear mass density 4.0 × 10⁻³ kg/m is under a tension of 90 N. A student wants the fundamental frequency to be 220 Hz but measures something different. The student has three possible adjustments: (1) increase tension, (2) use a thicker string of the same length and tension, or (3) press the string at a fret to shorten its vibrating length. The student argues that 'making the string heavier will make it vibrate faster because heavier things carry more energy.' The student also believes that shortening the string raises the pitch because 'shorter strings have bigger waves.' Assume both ends are fixed nodes.
1. Compute the current fundamental frequency, then determine what tension would be needed to reach 220 Hz with the original string.
Read the worked answer
Wave speed on the string: v = √(T/μ) = √(90 / 4.0×10⁻³) = √(22,500) = 150 m/s.
Fundamental (both ends fixed): λ = 2L = 2(0.65) = 1.30 m. f = v/λ = 150/1.30 ≈ 115 Hz.
To reach 220 Hz with the same string and length, f ∝ v ∝ √T, so: f₂/f₁ = √(T₂/T₁) → 220/115 = √(T₂/90). (1.913)² = T₂/90 → 3.66 = T₂/90 → T₂ ≈ 329 N.
So tension must rise from 90 N to about 330 N - a factor of about 3.7 in tension for a factor of about 1.9 in frequency, because frequency scales with the square root of tension.
2. Evaluate the student's two claims about heavier strings and shorter strings, using the wave relationship for a fixed-fixed string.
Read the worked answer
Heavier string claim: Incorrect. v = √(T/μ), so increasing μ decreases wave speed and therefore decreases frequency (f = v/2L). A thicker/heavier string at the same tension and length vibrates slower and sounds lower. 'More energy' does not mean higher frequency; frequency is set by speed and wavelength, not by stored energy.
Shorter string claim: The conclusion (higher pitch) is right, but the reasoning is wrong. Pressing a fret reduces L, and since λ = 2L, a shorter L means a shorter fundamental wavelength. With v fixed by tension and density, f = v/(2L) increases as L decreases. The student's phrase 'bigger waves' is backwards - shortening the string makes smaller wavelengths, which is why the frequency rises. The correct chain is: shorter L → shorter λ → higher f (at constant v).