Open organ pipe what does 6 beats per second mean?

[yjk91]

Homework Statement

You are playing a note that has a fundamental frequency of 428 Hz on a guitar string of length 56 cm. At the same time, your friend plays a fundamental note on an open organ pipe, and 6 beats per seconds are heard. The mass per unit length of the string is 2 g/m. Assume the velocity of sound is 343 m/s.

(a) What are the possible frequencies of the open organ pipe? (Enter your answers from smallest to largest.)

(b)What is the length of the organ pipe?

The Attempt at a Solution

What does 6 beats per seconds mean?
does it mean that there are 6 nodes?

 

[Ouabache]

What does 6 beats per seconds mean?

What happens when you play two notes close in frequency, notes that are almost but not equal in frequency?

 

[yjk91]

beats?

 

[Ouabache]

Correct, also known as beat frequency. That is what you are hearing at 6 bps. To hear what is going on, you could listen to the 2nd (low frequency) audio example under http://en.wikipedia.org/wiki/Beat_%28acoustics%29" , towards the bottom of that page.
To understand why you are hearing beats (and the mathematical relationship to the two contributing notes), you may want to look at http://hyperphysics.phy-astr.gsu.edu/hbase/sound/beat.html" .

 

[rwisz]

I would define beats per second as the perceived pulsating sound that occurs when two sound waves with slightly different frequencies interfere with each other. In this context, it means that the note played on the guitar string and the note played on the open organ pipe have frequencies that are close enough to create a beat frequency of 6 beats per second.

(a) The possible frequencies of the open organ pipe can be calculated using the formula f = nV/2L, where n is the harmonic number, V is the velocity of sound (343 m/s), and L is the length of the pipe. Since we know the fundamental frequency of the guitar string (428 Hz), we can set n=1 and solve for L to find the possible lengths of the open organ pipe. This gives us L = nV/2f = (1)(343 m/s)/(2*428 Hz) = 0.4 m. Therefore, the possible frequencies of the open organ pipe are 343/0.4 = 857.5 Hz, 1715 Hz, 2572.5 Hz, and so on.

(b) The length of the organ pipe is 0.4 m, as calculated in part (a).