76.3k views
1 vote
An organ pipe has length l. Its fundamental (i.e., lowest) frequency is proportional to

a. 1/l²
b. l
c. 1/l
d. l²

User Dguan
by
8.5k points

1 Answer

1 vote

Final answer:

The fundamental frequency of an organ pipe is inversely proportional to its length (1/l). Closed-pipe resonators have a fundamental wavelength of 4L, while open-pipe resonators have a fundamental wavelength of 2L. Both types exhibit a fundamental frequency that is dependent on the speed of sound, which varies with temperature.

Step-by-step explanation:

The fundamental (lowest) frequency of an organ pipe is proportional to 1/l, where 'l' is the length of the pipe. This relationship comes from the fact that, for an organ pipe that is closed at one end (also called a closed-pipe resonator), the fundamental wavelength (λ) is four times the length of the pipe (λ = 4L). The frequency (f) is calculated by dividing the speed of sound (v) by the wavelength (f = v/λ), leading to f = v/(4L), which simplifies to f being proportional to 1/L. For an organ pipe that is open at both ends, the fundamental wavelength is twice the length of the pipe (λ = 2L). Hence, the fundamental frequency is f = v/(2L), again showing that the frequency is inversely proportional to the length of the pipe. Closed-pipe resonator: If an organ pipe closed at one end produces a fundamental frequency of 256 Hz when air temperature is 18.0°C, we can find the length of the pipe using the relationship f = v/(4L). The speed of sound varies slightly with temperature, which would also affect the fundamental frequency. Open-pipe resonator: For a pipe open at both ends, like a flute or an oboe, the fundamental frequency is achieved when the standing wave inside the pipe has a wavelength of twice the length of the pipe.

User Yuval Adam
by
8.8k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.