You and a friend frequently play a trombone duet in a jazz band. During such performances it is critical that the two instruments be perfectly tuned. Since you take better care of your trombone, you decide to use your instrument as the standard. When you produce a tone that is known to be 470 Hz and your friend attempts to play the same note, you hear 4 beats every 3.00 seconds. Your ear is good enough to detect that your trombone is at a higher frequency. Determine the frequency of your friend's trombone. (Enter your answer to at least 1 decimal place.)

f₂ = 468.67 Hz

Explanation:

A beat is a sudden increase and decrease of sound. The beats are produced through the interference of two sound waves of slightly different frequencies. Now we have the following data:

The higher frequency tone = f₁ = 470 Hz

No. of beats = n = 4 beats

Time period = t = 3 s

The lower frequency note = Frequency of Friend's Trombone = f₂ = ?

Beat Frequency = fb

So, the formula for beats per second or beat frequency is given as:

fb = n/t

fb = 4 beats/ 3 s

fb = 1.33 Hz

Another formula for beat frequency is:

fb = f₁ - f₂

f₂ =  f₁ - fb

f₂ = 470 Hz - 1.33 Hz

f₂ = 468.67 Hz

its c i just checked it

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