Answer to Question #103306 in Mechanics | Relativity for Umar

Question

A typical mountaineer rope about 35cm long will give a distance of about 1.6m under the weight of a climber hanging freely on the load. Assuming the climbers mass is 80kg, estimate the period of vertical oscillation of the climber when dangling freely on the end of the rope how does this compare with the period of solving of the climber "pendulum"

Answer 1

A rope can be considered as a spring. Its length is related to the mass of the climber hanging freely on the load according to the formula

(1) $k\cdot \Delta x=mg$ ,

where $\Delta x= x_2-x_1=$1.6m-35cm= 1.6m-0.35m=1.25m, $m=80kg$ and $g=9.8ms^{-2}$acceleration of gravity. We can determine

(2) $k=\frac {mg}{\Delta x}=\frac{80\cdot 9.8}{1.25}=627.2Nm^{-1}$ .

The formula for the free oscillation period of a spring oscillator is known

(3) $T=2 \pi \sqrt{\frac{m}{k}}=6.28\cdot \sqrt{\frac{80kg}{627.2Nm^{-1}}}=$ 2.24 s

It was possible to substitute (2) in (3).

(4) $T=2 \pi \sqrt{\frac{m}{mg/\Delta x}}=2 \pi \sqrt{\frac{\Delta x}{g}}$

This formula (4) coincides with the formula for the small transverse oscillation of a mathematical pendulum, which is a point mass suspended on a massless rod or rope. One can see that usually, the transverse vibrations of the climber on the safety rope are much slower than the longitudinal ones, because usually the length of safety line is much longer then $\Delta x=1.25m$ .

Answer: The period of vertical oscillation of the climber when dangling freely on the end of the rope is 2.24 s, this is mush lower compare with the period of oscillation of the climber "pendulum".