Question

A uniform rope of length L and mass m is held at one end and whirled in a horizontal circle with angular velocity $\omega$. Ignore gravity. Find the time required for a transverse wave to travel from one end of the rope to the other.

Answer 1

$\frac{\pi}{\sqrt{2}\omega}$

Answer 2

1. Calculate the linear mass density $\mu = m/L$.

2. Determine the tension $T(x)$ at a distance $x$ from the held end by considering the centripetal force required for the part of the rope from $x$ to $L$. This yields $T(x) = \frac{1}{2} \mu \omega^2 (L^2 - x^2)$.

3. Calculate the speed of the transverse wave $v(x) = \sqrt{T(x)/\mu} = \omega \sqrt{(L^2 - x^2)/2}$.

4. The time taken to travel from $x=0$ to $x=L$ is $t = \int_0^L \frac{dx}{v(x)}$.

5. Evaluate the integral $t = \int_0^L \frac{\sqrt{2}}{\omega \sqrt{L^2 - x^2}} dx = \frac{\sqrt{2}}{\omega} [\arcsin(x/L)]_0^L = \frac{\sqrt{2}}{\omega} (\pi/2 - 0) = \frac{\pi}{\sqrt{2}\omega}$.

Answer: $\frac{\pi}{\sqrt{2}\omega}$