Question

A 2.31-kg rope is stretched between supports that are 10.4 m apart, and has a tension in it of 49.2 N. If one end of the rope is slightly tweaked, how long will it take for the resulting disturbance to reach the other end?

Answer 1

Answer:

The resulting disturbance takes approximately 0.70 seconds to reach the other end.

Explanation:

The velocity of a disturbance on a rope depends on the physical properties of the material and the tension exerted to it, the equation is:

$v=\sqrt{\frac{T}{\mu}}$  (1)

With v the velocity of the disturbance, T the tension of the rope and μ the linear density of the rope. The linear density of the rope is the relation between mass M and length L so we have:

$\frac{M}{L}$ (2)

Using (2) on (1):

$v=\sqrt{\frac{TL}{M}}=\sqrt{\frac{49.2*10.4}{2.31}}\simeq14.9\frac{m}{s}$ (3)

If we assume the velocity of the disturbance is constant, it will take for the resulting disturbance to reach the other ends a time t:

$t=\frac{L}{v}=\frac{10.4}{14.9}\simeq0.70\,s$ (4)