Answer:
It is producing either a 435-Hz sound or a 441-Hz sound.
Step-by-step explanation:
When two sound of slightly different frequencies interfere constructively with each other, the resultant wave has a frequency (called beat frequency) which is equal to the absolute value of the difference between the individual frequencies:
(1)
In this problem, we know that:
- The frequency of the first trombone is
![f_1 = 438 Hz](https://img.qammunity.org/2020/formulas/physics/high-school/8mfbr4gsasknd87b6g69d7fwwreixfp3z5.png)
- 6 beats are heard every 2 seconds, so the beat frequency is
![f_B=(6)/(2 s)=3 Hz](https://img.qammunity.org/2020/formulas/physics/high-school/t7wa8poarng55j5of5lua3k6j3qdg0zgs8.png)
If we insert this data into eq.(1), we have two possible solutions for the frequency of the second trombone:
![f_2 = f_1 - f_B = 438 Hz - 3 Hz = 435 Hz\\f_2 = f_1+f_B = 438 Hz+3 Hz=441 Hz](https://img.qammunity.org/2020/formulas/physics/high-school/4ngybazwy3tbvpd604cx0cd0nyg0fvmqv7.png)