Questions: Acoustic Resonance in Strings and Tension

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A guitar string of length L resonates at fundamental frequency f₁. A guitarist frets the string at the midpoint, halving the vibrating length while keeping tension and mass density unchanged. The new fundamental frequency is...

A2f₁ — halving the length doubles the fundamental frequency
Bf₁/2 — a shorter string vibrates more slowly
C4f₁ — frequency scales as 1/L²
Df₁√2 — frequency scales as 1/√L
Question 2 Multiple Choice

Why are the resonant frequencies of a fixed string spaced at integer multiples of the fundamental (f₁, 2f₁, 3f₁, ...)?

ABecause each harmonic requires an integer number of half-wavelengths to fit between the fixed endpoints, and frequency is inversely proportional to wavelength at fixed wave speed
BBecause string tension increases proportionally with each successive harmonic, raising the frequency by equal steps
CBecause higher harmonics travel faster through the string, raising their frequency above the fundamental
DBecause each harmonic corresponds to a different linear mass density along the string
Question 3 True / False

Increasing the tension of a guitar string raises all of its resonant frequencies — fundamental and harmonics alike.

TTrue
FFalse
Question 4 True / False

The second harmonic of a string has twice the wavelength of the fundamental.

TTrue
FFalse
Question 5 Short Answer

Explain why a string fixed at both ends cannot resonate at an arbitrary driving frequency — why only certain discrete frequencies produce standing waves.

Think about your answer, then reveal below.