Answer to Question #132335 in Mechanics | Relativity for Joseph Baroku

Question

A bungee jumper with mass of 65kg jumps from an edge of a cliff. After reaching home lowest point, he oscillates up and down, hitting the low point two more times in 43 seconds. He comes to rest 25m below the starting point. Estimated the spring stiffness constant k and the un-stretched length of the bungee cord assuming motion is simple harmonic.

Answer 1

solution:

mass of jumper(m)=65kg

frequency of oscilation

$f=\frac{2}{43}=0.046Hz$

angular frequency

$\omega=2\pi f=2\times3.14\times0.046=0.288rad/s$

stiffness constant

$k=m\omega^2=65\times(0.288)^2=5.39N/m$

in final position

$kx=mg$

$x=\frac{mg}k=\frac{65\times9.8}{5.39}=118.30m$

therefore

$25=unstretched\space length+118.30$

unstretched length =25-118.30=-93.3m