String Resonance, Harmonics, and Tension Adjustment
A string of length 1.20 m and linear density 2.5 g/m is stretched between two fixed supports. The tension is set to 90 N. A small vibration generator attached near one end drives the string, and the full length vibrates; no movable bridge is used. A tuning fork of frequency 316 Hz is sounded nearby, and the string is observed to vibrate strongly when the tension is adjusted slightly. A student claims the string must be in its fundamental mode because that is the loudest mode, and proposes increasing tension to raise the pitch. You are asked to identify the mode and the tension required for resonance with the fork.
1. Calculate the fundamental frequency and determine which harmonic is close to 316 Hz.
Read the worked answer
For a string fixed at both ends, f_n = (n/2L)√(T/μ). Here L = 1.20 m, μ = 2.5×10^-3 kg/m, and T = 90 N. The wave speed is v = √(T/μ) = √(90/2.5×10^-3) = √36000 = 189.7 m/s. Therefore the fundamental is f_1 = v/(2L) = 189.7/2.40 = 79.1 Hz. The harmonics are integer multiples: 79.1, 158.1, 237.2, 316.2 Hz. The tuning fork at 316 Hz is therefore close to the fourth harmonic (n = 4), not the fundamental. The string has four loops and five nodes, including the two fixed ends.
2. Calculate the tension needed for the string to resonate exactly at 316 Hz in that mode, and state whether the student should increase or decrease the current tension.
Read the worked answer
Use f_n = (n/2L)√(T/μ) and solve for T at n = 4 and f = 316 Hz. Rearranging gives T = μ(2Lf/n)^2. The bracket is 2×1.20×316/4 = 0.600×316 = 189.6 m/s, so T = 2.5×10^-3 × (189.6)^2 = 2.5×10^-3 × 3.595×10^4 = 89.9 N. The present tension is 90 N, so the student should decrease the tension very slightly. Since f is proportional to √T, a small reduction in T lowers the frequency; the needed change is less than 0.2 N, which explains why the resonance appeared when the tension was adjusted slightly.
3. Evaluate the student's claim that the strong vibration must be the fundamental because that mode is loudest.
Read the worked answer
The observation is not evidence for the fundamental. The string is driven by the nearby tuning fork at 316 Hz, not by a broadband impulse; it responds strongly when one of its normal-mode frequencies is close to the driving frequency. The fundamental at 79 Hz would respond if the fork were near 79 Hz, but it is not. The fourth harmonic has four antinodes, so it can be excited with large amplitude even though it is not the lowest mode. The student is also conflating 'loudest' with 'lowest frequency'; the amplitude at resonance depends on driving strength and damping, not simply on the mode number. The correct conclusion is that the string is in the n = 4 mode.