Answer
(a) $m = 1.2~kg$
(b) $m = 0.29~kg$
(c) $m = 0.046~kg$
Work Step by Step
(a) We can find the speed required to produce one loop as:
$v = (f)(2L)$
$v = (60~Hz)(2)(1.5~m)$
$v = 180~m/s$
We can find the required tension which results in this speed.
$v = \sqrt{\frac{F_T}{\mu}}$
$F_T = v^2~\mu$
$F_T = (180~m/s)^2(3.5\times 10^{-4}~kg/m)$
$F_T = 11.34~N$
We can find the required mass.
$m = \frac{F_T}{g}$
$m = \frac{11.34~N}{9.80~m/s^2}$
$m = 1.2~kg$
(b) We can find the speed required to produce two loops.
$v = (f)(L)$
$v = (60~Hz)(1.5~m)$
$v = 90~m/s$
We can find the required tension which results in this speed.
$v = \sqrt{\frac{F_T}{\mu}}$
$F_T = v^2~\mu$
$F_T = (90~m/s)^2(3.5\times 10^{-4}~kg/m)$
$F_T = 2.835~N$
We can find the required mass.
$m = \frac{F_T}{g}$
$m = \frac{2.835~N}{9.80~m/s^2}$
$m = 0.29~kg$
(c) We can find the speed required to produce five loops.
$v = (f)(\frac{2L}{5})$
$v = (60~Hz)\frac{(2)(1.5~m)}{5}$
$v = 36~m/s$
We can find the required tension which results in this speed.
$v = \sqrt{\frac{F_T}{\mu}}$
$F_T = v^2~\mu$
$F_T = (36~m/s)^2(3.5\times 10^{-4}~kg/m)$
$F_T = 0.4536~N$
We can find the required mass.
$m = \frac{F_T}{g}$
$m = \frac{0.4536~N}{9.80~m/s^2}$
$m = 0.046~kg$
