What is the time taken by a pulse generated at the bottom of a string to reach the top?

[hdje]

Homework Statement

In a test the time taken by a pulse (generated at the bottom) to reach the top of the string was asked. Mass per unit length was 0.001kg/m and length was 1m.

Relevant Equations

v = √(T/μ)

I have been taught that speed of pulse is v = √(T/μ), but here tension varies at each point therefore I cannot just simply apply the formula. I think integration would be needed, I tried but ended up nowhere. Can someone help me find the time?

 

[nasu]

Can you post the text of the question?

 

[hdje]

Can you post the text of the question?

A string of length 1 m and mass per unit length o.oo1kg/m is hanging vertically. A small pulse is generated at its lower end. The pulse reaches the top end in approximately: (g=10m/s^2)

(1) 1/sqrt(10) s
(2) 3/sqrt(10) s
(3) 2/sqrt(10) s
(4) sqrt(5)/2

 

[nasu]

As the tension is not given, I would assume that you need to take it as produced by the string's own weight. It will increase linearly from zero at the bottom to the full weight at the top.

 

[haruspex]

I think integration would be needed

So how long would it take the pulse to get from a point distance y from the bottom to a point y+dy from the bottom?

 

[nasu]

You could solve it without having to do an integral if you find out the relationship between pulse velocity at position y and the distance travelled by the pulse up to that position. You may be familiar with such a relationship from previous study of motion.

 

[hdje]

I think I got it, but with integration. Correct me if I'm wrong.

hence the answer will be (3) 2/sqrt(10) s

 

[haruspex]

I think I got it, but with integration. Correct me if I'm wrong.

View attachment 342144
View attachment 342146
hence the answer will be (3) 2/sqrt(10) s

Looks right.

 

[nasu]

It is v and not dv in your first formula:
It is v=dy/dt and not dv.
But the answer is right.

Once you see that the square of velocity is proportional to the distance travelled you may remember the same relationship holds for uniform accelerated motion with zero initial velocity. So the average velocity is half the final velocity. And the time is the length of the string divided by the average velocity.