Question
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
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
Related Questions
Two identical violin strings, when i tune and stretched with the same tension, have a fundamental fr...
Two identical mandolin strings under 200 N of tension are sounding tones with frequencies of 577 Hz....
What is the BEST description of a chord?
the numbered order of the guitar strings from top to b...