Answer
(a) $f = 33.3~Hz$
(b) The required tension is $300~N$
Work Step by Step
(a) We can find the speed of the wave in the cord:
$v = \sqrt{\frac{F}{\mu}}$
$v = \sqrt{\frac{12~N}{1.2\times 10^{-3}~kg/m}}$
$v = 100~m/s$
We can find the fundamental frequency:
$f = \frac{v}{\lambda}$
$f = \frac{v}{2L}$
$f = \frac{100~m/s}{(2)(1.5~m)}$
$f = 33.3~Hz$
(b) We can find the required wave speed in the cord:
$v = \lambda_3~f_3$
$v = (\frac{2L}{3})(f_3)$
$v = (\frac{(2)(1.5~m)}{3})(0.50\times 10^3~Hz)$
$v = 500~m/s$
We can find the required tension:
$\sqrt{\frac{F}{\mu}} = v$
$\frac{F}{\mu} = v^2$
$F =\mu~ v^2$
$F =(1.2\times 10^{-3}~kg/m)(500~m/s)^2$
$F = 300~N$
The required tension is $300~N$
