It is stated in the previous problem that a pilse travels from the bottom to the top of a hanging rope of length L in the time interval `Delta=2sqrt(L/g)`. Use this result to answer the following question. (It is not neccesary to set up any new integrations.) (a) over what time interval does a pulse travel halfway up the rope? Give your answer as a fraction of the quantity `2sqrt(L/g)`. (b) A pulse starts travelling up the rope. how far it travelled after a time interval `sqrtL/g)`?

Explanation

The wave pulse travels faster as it goes up the rope because the tension higer in the rope is greater (to support the weight of the rope below it ). Therefore it should take mote than half the total time `Deltat` for the wave to travel halfway up the rope. likewise, the pulse should travel less than halfway up the rope in time `Deltat//2`.
By using the time relationship given in the problem and making suitable substitution, we can find the required time and distance.
(a) From the equation given, the time for a pulse to pulse to travel any distance, d, up from the bottom of the rope is `Deltat_(d)=2sqrt(d)//(g)`. so the time for a pulse to travel a distance `L//2` from the botom is `Deltat_(1//2)=2sqrt(L)/(2g)=0.707(2sqrt(L)/(g))`
(b) Likewise, the distance a pulse a pulse travels from the bottom of a rope in a time `Deltat_(d)` is `d=(gDeltat_(d)^(2)`. so the distance travelled by a pulse after a time
`Delta_(d)=sqrt(L)/(g) is d =(g(L//g))/(4)=(L)/(4)`