You are playing a note that has a fundamental frequency of 440.00 Hz on a guitar string of length 59.1 cm. At the same time, your friend plays a fundamental note on an open organ pipe, and 4.0 beats per seconds are heard. The mass per unit length of the string is 1.93 g/m. Assume the velocity of sound is 343 m/s.
a) When the guitar string is tightened, the beat frequency decreases. Find the original tension in the string.
b)What is the length of the organ pipe?
a. well, beats are result of either a freq of 440+-4, but it is low, so f=436hz
you know half wavelength = 59.1cm
you know mass per length. You do not know wave velocity on the string.
Use http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html
solve for wave velocity first:
wavevelocity=2*length*frequency
b. Pipe organ, open pipe, lambda=2*length
you know frequency 440hz, solve for wavlength, lambda=speedsoundinAir/freq
now knowing wavelength, solve for length (lambda/2)
a) To find the original tension in the guitar string, we need to calculate the beat frequency. The beat frequency is the difference between the frequencies of the two notes being played. In this case, the beat frequency is given as 4.0 beats per second.
Let's assume the frequency of the note played on the organ pipe is f1. The beat frequency can be calculated using the formula:
Beat frequency = |f1 - 440.00 Hz|
Since the beat frequency is positive, we can conclude that the frequency of the note played on the organ pipe (f1) is greater than 440.00 Hz.
Now, we need to calculate the frequency of the note played on the organ pipe. Using the formula for the fundamental frequency of an open organ pipe:
f1 = (v/2L),
where v is the velocity of sound (343 m/s) and L is the length of the organ pipe.
Let's assume the length of the organ pipe is L1. Putting it all together, we have:
f1 = (343 m/s) / (2 * L1).
Since we know the beat frequency is 4.0 beats per second and the frequency of the note played on the guitar string is 440.00 Hz, we can write:
f1 - 440.00 Hz = 4.0 Hz.
Now we have two equations:
f1 = (343 m/s) / (2 * L1) (equation 1)
f1 - 440.00 Hz = 4.0 Hz (equation 2)
Solving equation 2 for f1:
f1 = 444.00 Hz.
Substituting this into equation 1:
444.00 Hz = (343 m/s) / (2 * L1).
Rearranging the equation:
2 * L1 = (343 m/s) / 444.00 Hz,
L1 = (343 m/s) / (444.00 Hz * 2).
Now we can calculate the length of the organ pipe, L1.
b) Substituting the given values into the equation:
L1 = (343 m/s) / (444.00 Hz * 2).
L1 ≈ 0.386 m.
So, the length of the organ pipe is approximately 0.386 meters.