## College Physics (4th Edition)

A weight of $1284~N$ should be hung from the wire.
We can find the speed of sound in air: $v = 331+0.6~T = 331+(0.6)(18.0^{\circ}C) = 341.8~m/s$ We can find the fundamental frequency of the tube: $f = \frac{v}{\lambda}$ $f = \frac{v}{4L}$ $f = \frac{341.8~m/s}{(4)(1.50~m)}$ $f = 57.0~Hz$ We can find the required wave speed in the wire: $v = \lambda~f$ $v = 2L~f$ $v = (2)(1.0~m)(57.0~Hz)$ $v = 114~m/s$ We can find the mass of 1.0-m of the wire: $m = V~\rho$ $m = A~L~\rho$ $m = \pi~r^2~L~\rho$ $m = (\pi)~(2.00\times 10^{-3}~m)^2~(1.0~m)~(7860~kg/m^3)$ $m = 0.0988~kg$ We can find the required tension in the wire: $\sqrt{\frac{F}{m/L}} = v$ $\sqrt{\frac{F~L}{m}} = v$ $\frac{F~L}{m} = v^2$ $F = \frac{m~v^2}{L}$ $F = \frac{(0.0988~kg)(114~m/s)^2}{1.0~m}$ $F = 1284~N$ A weight of $1284~N$ should be hung from the wire.