Answer
(a) v' = 283 cm/s
(b) v' = 100 cm/s
(c) v' = 400 cm/s
(d) v' = 200 cm/s
Work Step by Step
We can write an expression for the original speed of the wave which is 200 cm/s:
$v = \sqrt{\frac{F_T}{\mu}}$
$v = \sqrt{\frac{F_T}{(m/L)}}$
$v = \sqrt{\frac{F_T~L}{m}}$
(a) We can find the new speed of the wave:
$v' = \sqrt{\frac{2F_T~L}{m}}$
$v' = \sqrt{2}~\sqrt{\frac{F_T~L}{m}}$
$v' = \sqrt{2}~(200~cm/s)$
$v' = 283~cm/s$
(b) We can find the new speed of the wave:
$v' = \sqrt{\frac{F_T~L}{4m}}$
$v' = \sqrt{\frac{1}{4}}~\sqrt{\frac{F_T~L}{m}}$
$v' = \frac{1}{2}~\sqrt{\frac{F_T~L}{m}}$
$v' = \frac{1}{2}~(200~cm/s)$
$v' = 100~cm/s$
(c) We can find the new speed of the wave:
$v' = \sqrt{\frac{F_T~(4L)}{m}}$
$v' = \sqrt{4}~\sqrt{\frac{F_T~L}{m}}$
$v' = 2~\sqrt{\frac{F_T~L}{m}}$
$v' = (2)~(200~cm/s)$
$v' = 400~cm/s$
(d) We can find the speed of the wave:
$v' = \sqrt{\frac{F_T~(4L)}{4m}}$
$v' = \sqrt{\frac{F_T~L}{m}}$
$v' = 200~cm/s$