Help with a physics sound problem.

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Two identical strings on cellos are tuned to 440 Hz, but one string's tension decreases by 1.5% due to a slipped peg. The calculated frequency of the second string becomes 433.4 Hz, leading to a beat frequency of 6.6 Hz. However, the correct beat frequency is 3.3 Hz, indicating a missing formula in the initial calculations. The discussion highlights the importance of understanding the relationship between tension, wave velocity, and pitch. Ultimately, the correct approach led to finding the right answer.
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1. Two identical strings on different cellos are tuned to the 440-Hz A note. The peg holding one of the strings slips, so its tension is decreased by 1.5%. What is the beat frequency heard when the strings are played together?
f₁=440Hz; Tension decrease=1.5%2. f(beat)=|f₁-f₂|3.
f₂=440Hz - (440Hz*.015)=433.4Hz
f(beat)=|440Hz-433.4Hz|=6.6Hz

The answer is supposed to be 3.3Hz. I would really appreciate the help. Thanks.
 
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I think you're missing a key formula here.
Remember that the velocity of the wave on a tensioned string is

v = √(F / µ) where µ is the string density. Use it more conceptually now, and think about the relationship of the force to the velocity, and the velocity to the pitch and wavelength.
 
Thank you very much. I got the right answer now. Much appreciated. :)
 
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