a 60.0 kilogram bungee jumper hangs from the end of her bungee rope. if she oscillates up and down with a period of 8.00 seconds, what is the spring constant (k) of her bungee rope? [note - the units of k are n/m

Respuesta :

The spring constant of bungee rope is 36.98n/m.

The spring constant, k, is a measure of the stiffness of the spring. It is measured by using Hooke’s law, it is the force required to compress or extend a spring is proportional to the length stretched.

This restoring force follows Hooke’s Law, which relates spring force to constant spring force.

Spring Force = -(Spring Constant) × (Displacement)

F=-kx

F: The restoring force of the spring, directed towards equilibrium.

K: The spring constant in N.m-1.

X: The spring’s displacement from its equilibrium position.

Given

m=60.kilogram

Time period T=8.00s

For this, we have a formula,

                 

                     [tex]T=2\pi *\sqrt{\frac{m}{k} } \\\\8=2*3.14*\sqrt{\frac{60}{k} } \\squaring on both sides we get\\\\64=\frac{39.4384*60}{k} \\\\k=\frac{2366.304}{64} \\\\k=36.98[/tex]

∴ The spring constant(k)=36.98n/m

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