Il A 40-cm- long tube has a 40-cm- long insert that can be...

Il A $40-\mathrm{cm}-$ long tube has a $40-\mathrm{cm}-$ long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. As the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length $L$ is $42.5 \mathrm{cm}, 56.7 \mathrm{cm},$ and $70.9 \mathrm{cm} .$ What is the frequency of the tuning fork? Assume $v_{\text {sound }}=343 \mathrm{m} / \mathrm{s}$.

Key Concepts

Standing Waves

Standing waves occur when two waves of the same frequency and amplitude traveling in opposite directions interfere, resulting in fixed points of no motion (nodes) and points of maximum displacement (antinodes). This concept is fundamental in understanding how sound waves resonate inside tubes, where the specific arrangement of nodes and antinodes gives rise to resonant frequencies.

Resonance in Air Columns

Resonance in air columns happens when the physical dimensions of a tube correspond to an integral multiple of half wavelengths of a sound wave. This matching condition causes the sound wave to reinforce itself, creating distinct resonant modes. The exact resonant lengths depend on the boundary conditions (such as open or closed ends) of the tube.

Frequency-Wavelength-Speed Relationship

The frequency of a sound is intrinsically linked to its wavelength and the speed of sound by the relation f = v/?. This principle allows one to calculate the frequency of a sound source if the wavelength (derived from the resonant conditions in the tube) and the speed of sound in air are known.

Mode Differences and Harmonics

Differences between successive resonant lengths in an air column are typically equal to half a wavelength, indicating the transition from one harmonic mode to the next. Recognizing these incremental differences enables one to determine the wavelength of the standing wave and thus calculate the corresponding frequency of the sound.

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