Asked by Jeanette
Two identical guitar strings under 108 N of tension sound tones with frequencies of 528 Hz. The peg of one string slips slightly, and the tension in it drops to 91.0 N. How many beats per second are heard?
Answers
Answered by
bobpursley
v= sqrt(T/mu)= freq*wavelength
so wavelength remais the same, so
(108/mu / 91/mu)=(528/f)^2
f^2=528^2/(108/91)
f= 528 (sqrt(91/108) about 485 (work it out)
beats= 528-485
so wavelength remais the same, so
(108/mu / 91/mu)=(528/f)^2
f^2=528^2/(108/91)
f= 528 (sqrt(91/108) about 485 (work it out)
beats= 528-485
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