Answer
$k = 68.0~N/m$
The unstretched length of the cord is 15.6 meters.
Work Step by Step
We can find the period as:
$T = \frac{time}{oscillations}$
$T = \frac{43.0~s}{7~oscillations}$
$T = 6.143~s$
We then find the spring constant as:
$T = 2\pi~\sqrt{\frac{m}{k}}$
$k = \frac{(2\pi)^2~m}{T^2}$
$k = \frac{(2\pi)^2(65.0~kg)}{(6.143~s)^2}$
$k = 68.0~N/m$
We then find the distance $x$ that the cord is stretched;
$kx = mg$
$x = \frac{mg}{k}$
$x = \frac{(65.0~kg)(9.80~m/s^2)}{68.0~N/m}$
$x = 9.37~m$
Since the person comes to rest 25.0 meters below the bridge, we can find the unstretched length of the cord.
$L = 25.0~m-9.37~m = 15.6~m$
