Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note A should have the frequency 440 Hz} and the note E should be at 659 Hz}. The tuner can determine this by listening to the beats between the third harmonic of the A and the second harmonic of the E.
Part A: A tuner first tunes the A string very precisely by matching it to a 440 Hz tuning fork. She then strikes the A and E strings simultaneously and listens for beats between the harmonics. What beat frequency indicates that the E string is properly tuned?'
Part B: The tuner starts with the tension in the E string a little low, then tightens it. What is the frequency of the E string when she hears four beats per second?
these answers are not correct
ahah yeah bobpursley seems like you had difficulty yourself
actually bob, it's 658. You subtract freq E by one harmonic 659-1=658
Part A:
To determine the beat frequency that indicates the E string is properly tuned, we need to understand the relationship between the frequency of the A string, the frequency of the E string, and the beats.
The A string has a frequency of 440 Hz.
The E string has a frequency of 659 Hz.
The tuner listens for beats between the third harmonic of the A string and the second harmonic of the E string. The harmonic of a string refers to a multiple of its fundamental frequency.
The third harmonic of the A string would have a frequency of 3 times the fundamental frequency, which is 3 * 440 Hz = 1320 Hz.
The second harmonic of the E string would have a frequency of 2 times the fundamental frequency, which is 2 * 659 Hz = 1318 Hz.
The beat frequency is the difference between these two frequencies. Therefore, the beat frequency can be calculated as 1320 Hz - 1318 Hz = 2 Hz.
Therefore, when the E string is properly tuned, the tuner would hear a beat frequency of 2 Hz.
Part B:
In this part, we need to find the frequency of the E string when the tuner hears four beats per second.
Knowing that the beat frequency is equal to the difference between the frequencies of the two strings, we can set up an equation.
Let's assume the frequency of the E string is x Hz.
The third harmonic of the A string is still 1320 Hz, as mentioned earlier.
The second harmonic of the E string would be 2x Hz.
Therefore, the beat frequency is given by 1320 Hz - 2x Hz.
Given that the tuner hears four beats per second, we can set up the following equation:
4 beats/second = 1320 Hz - 2x Hz.
Solving for x, we get:
2x Hz = 1320 Hz - 4 beats/second.
Simplifying further:
2x Hz = 1320 Hz - 4 Hz.
2x Hz = 1316 Hz.
Now dividing both sides of the equation by 2, we get:
x Hz = 658 Hz.
Therefore, when the tuner hears four beats per second, the frequency of the E string is 658 Hz.
beats=440(3)-659*2
b) 440*3-F(2)=4
What I wonder, is what could have possibly been your difficulty with this?