. Mass 1 has a momentum of 20 kg*m/s. Mass 2 has a momentum of 50 kg*m/s.
1 answer:
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(a) 2 Hz
The frequency of the nth-harmonic is given by
where
is the fundamental frequency
Therefore, the frequency of the third harmonic of the A () is
while the frequency of the second harmonic of the E () is
So the frequency difference is
(b) 2 Hz
The beat frequency between two harmonics of different frequencies f, f' is given by
In this case, when the strings are properly tuned, we have
- Frequency of the 3rd harmonic of A-note: 1320 Hz
- Frequency of the 2nd harmonic of E-note: 1318 Hz
So, the beat frequency should be (if the strings are properly tuned)
(c) 1324 Hz
The fundamental frequency on a string is proportional to the square root of the tension in the string:
this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.
In this situation, the beat frequency is 4 Hz (four beats per second):
And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,
and , we have two possible solutions for
:
However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be
1324 Hz