Question

A pulse is sent traveling along a rope under a tension of 36 N whose mass per unit length abruptly changes, from 14 kg/m to 49 kg/m. The length of the rope is 2.3 m for the first section and 2.6 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin?

Answer 1

We have the case as shown in the figure:

The pulses traveling along rope 1 and 2 have the same speed respectively with the following equation:

$v=\sqrt{\frac{T}{\mu }}$

$T=tension=36\:N$

$\mu =mass\:per\:unit\:length$

$\mu_1 =14\:\frac{kg}{m}$

$\mu_2 =49\:\frac{kg}{m}$

$v_1=\sqrt{\frac{36\:N}{14\:\frac{kg}{m}}}=1.6\:\frac{m}{s}$

$v_2=\sqrt{\frac{36\:N}{49\:\frac{kg}{m}}}=0.86\:\frac{m}{s}$

These are the speeds of the pulse along the rope 1 and 2 respectively. The speed does not depend in amplitud.

The time are:

$t_A=t_{1}=\left ( \frac{l_{1}}{v_1} \right )=\left ( \frac{2.3\:m}{1.6\:\frac{m}{s}} \right )=1.44\:s$

$t_B=2t_2+t_1$

This is the pulse from discontinuity to start point (rope 2 twice + rope 1) and getting back:

$t_3=t_1+2\left ( \frac{2.6\:m}{v_2} \right )=7.48\:s$

$t_4=t_3+\left ( \frac{2.3\:m}{v_1} \right )=8.92\:s$

$\Delta t=t_4-t_2={\color{Blue} 6.04\:s}$