Final answer:
In a closed-end tube, standing waves resonate at specific lengths corresponding to odd multiples of a quarter wavelength. The specified lengths for resonance can be calculated considering the frequency of the sound, the speed of sound in air, and the sequence of odd harmonics.
Step-by-step explanation:
The question pertains to the concept of sound resonating within a closed-end air column. When a sound wave of frequency 300 Hz resonates in a tube with a length of 2 m that is closed on one end, the length at which resonance occurs is determined by the properties of the standing wave that can form in such a tube.
In a closed-end tube, the standing waves form such that there is always a node at the closed end and an antinode at the open end. The fundamental frequency (first harmonic) corresponds to the tube length being one-quarter the wavelength (λ/4).
For the second resonance (third harmonic), the tube length accommodates three-quarters of a wavelength (3λ/4), and for the third resonance (fifth harmonic), it is 5λ/4 and so on, with only odd harmonics forming in a closed-end tube.
To find the lengths where resonance occurs for the first, second, and third cases, consider the following steps:
- First, calculate the wavelength (λ) of the sound wave using the speed of sound in air (assuming 343 m/s at room temperature) and the given frequency (f): λ = v/f.
- Next, find the length (L) at which each resonance occurs by using L = nλ/4, where n=1, 3, 5 for the first, second, and third resonances, respectively.”