Answer
(a) $f_1 = 408~Hz$
(b) The person could hear frequencies up to the 24th harmonic.
Work Step by Step
(a) We can find the speed of the wave.
$v = \sqrt{\frac{F_T}{\mu}}$
$v = \sqrt{\frac{F_T}{(m/L)}}$
$v = \sqrt{\frac{F_T~L}{m}}$
$v = \sqrt{\frac{(800~N)(0.400~m)}{0.00300~kg}}$
$v = 326.6~m/s$
We can find the fundamental frequency, which occurs when n = 1:
$f_1 = \frac{v}{2L}$
$f_1 = \frac{326.6~m/s}{(2)(0.400~m)}$
$f_1 = 408~Hz$
(b) We can find the harmonic $n$ which could be heard by a person who can hear frequencies up to 10,000 Hz.
$n~f_1 \leq 10,000~Hz$
$n \leq \frac{10,000~Hz}{f_1}$
$n \leq \frac{10,000~Hz}{408~Hz}$
$n \leq 24.5$
The person could hear frequencies up to the 24th harmonic.
