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A musician measures the frequencies of the audible standing waves in an organ pipe. She finds two adjacent tones at 532 and 684 Hz.

(a) On the basis of this discovery, the musician computes the pipe's fundamental frequency. What is its value (in Hz)?
Hz
(b) Is the pipe open at both ends or only one?
O open at both ends
O open at only one end
(c) The air within the pipe has a temperature of 20°C and is at atmospheric pressure. How long (in m) is the pipe?
m

User Rozuur
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1 Answer

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Answer: (a) The fundamental frequency of an organ pipe can be calculated using the formula:

f1 = v / (2L)

where f1 is the fundamental frequency, v is the speed of sound in air, and L is the length of the pipe.

Given that the musician measured two adjacent tones at 532 and 684 Hz, these correspond to the first harmonic (fundamental frequency) and the second harmonic (first overtone) respectively. The frequency of the second harmonic is twice the frequency of the fundamental frequency.

So, f1 = 532 Hz / 2 = 266 Hz

The fundamental frequency of the pipe is 266 Hz.

(b) Based on the given information, since the musician measured two adjacent tones, the frequencies of which are in the ratio of 1:2 (532 Hz and 684 Hz), this indicates that the pipe is open at both ends. In open-ended pipes, such as organ pipes, both ends are free to vibrate, allowing for the presence of odd harmonics (1st, 3rd, 5th, etc.) in addition to the even harmonics (2nd, 4th, 6th, etc.).

(c) To calculate the length of the pipe, we can use the formula:

L = v / (4f1)

where L is the length of the pipe, v is the speed of sound in air, and f1 is the fundamental frequency.

Given that the temperature is 20°C, the speed of sound in air at this temperature is approximately 343 m/s.

Plugging in the values, we get:

L = 343 m/s / (4 x 266 Hz) ≈ 0.808 m

So, the length of the pipe is approximately 0.808 meters.

User Logcat
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