## College Physics (4th Edition)

(a) $f = 33.3~Hz$ (b) The required tension is $300~N$
(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 $300~N$