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?

Answer 1

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