A bungee trampoline (such as you might see on Brighton sea-front in the summer)

Question

A bungee trampoline (such as you might see on Brighton sea-front in the summer) consists of two bungee ropes each attached to a supporting pole at one end and a harness at the other end. For the purposes of this question, assume that the harness is small and that the bungee ropes can be modelled as perfect, massless springs of spring constant k and with a combined natural length. 2d 4m, equal to the separation of the poles (upper, horizontal, dotted line in the figure). With only the harness, of mass 1 kg, attached to the bungee ropes, the bungee ropes hang at an angle of theta 1 = 20 degree to the horizontal (middle, dashed line in the figure). When a holiday-maker is strapped into the harness, the ropes then make an angle of theta 2 = 60 degree to the horizontal (lower, dash-dotted line in the figure). Assuming that the motion of the customer is confined to the vertical direction, show that the equation of motion is

Explanation / Answer

Note that the summation of forces along x (vertical direction) is

mg - kLsin(A) = mx''     [1]

Here, L is the change in length of the rope.

A = angle the rope makes with the horizontal.

Thus, sin A = x/sqrt(x^2 + d^2).

Note that

unstretched length (of each side) = d

stretched length = sqrt(d^2 + x^2) = d sqrt(1 + x^2/d^2)

Thus,

L = change in length = d - d sqrt(1 + x^2/d^2)

Factoring d,

L = d[1 - sqrt(1 + x^2/d^2)]

Thus, equation [1] becomes

mg - kd[1 - sqrt(1 + x^2/d^2)][x/dsqrt(1 + x^2/d^2)] = mx''     [1]

Dividing both sides by m,

x'' = g - kd/m [1 - sqrt(1 + x^2/d^2)][x/dsqrt(1 + x^2/d^2)]

Dividing both sides by d and simplifying,

x.. = g/d - k/m [1 - sqrt(1 + x^2)]   [DONE!]

where x.. = x''/d.

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