Question

A loose coil of flexible rope of length L and mass m rests on a frictionless table. A force S is applied at one end of the rope and more and more rope is pulled from the coil with a constant velocity v. A transverse pulse is produced in the rope while the rope is being pulled from the coil. The velocity of the pulse in the rope (relative to the surface of the table) will be (assuming the pulse moves in the direction of the pull.)

• A. zero
• B. 2v
• C. $\sqrt{\frac{SL}{m}}$
• D. $\frac{1}{2}\sqrt{\frac{SL}{m}}$

Answer 1

Answer: (B)

2v

Answer 2

The velocity of a transverse pulse in a rope relative to the rope is given by $v_{pulse, rope} = \sqrt{T/\mu}$, where T is the tension and $\mu$ is the linear mass density.

The rope is being pulled with a constant velocity v relative to the table.

The velocity of the pulse relative to the table is $v_{pulse, table} = v_{pulse, rope} + v_{rope, table}$. Assuming the pulse moves in the direction of the pull, $v_{pulse, table} = v_{pulse, rope} + v$.

For a rope being pulled from a coil with constant velocity v, the tension at the point where the rope leaves the coil is $T_0 = \mu v^2$.

The velocity of a pulse relative to the rope at this point is $\sqrt{T_0/\mu} = \sqrt{\mu v^2 / \mu} = v$.

The velocity of this pulse relative to the table is $v_{pulse, table} = v + v = 2v$.

This result, that the velocity of a pulse relative to the coil (table) is 2v, is a known result for this scenario.