# Vertical string problem

1. Mar 16, 2005

### skiboka33

A rope of mass m and length L is suspended vertically. Show that a transverse wave pulse will travel the length of the tope in a time t = 2(L/g)^1/2.

stuck... i tried:

v = (T/u)^1/2 ; where T = m(X/L)g and u = m/L

the X is the distance from the bottom of the string...

so i got v= (xg)^1/2

t= (L-X)/ (xg)^1/2

which i can't simplify to t = 2(L/g)^1/2.

Where did I go wrong here? logic is making sense to me but im no getting the answer. thanks.

2. Mar 16, 2005

### Staff: Mentor

You found v as of function of x correctly, but then just multiplied by the distance to find the time. Since v is not constant, you must integrate:
$$v = \frac{dx}{dt} = (xg)^{1/2}$$
$$t = \int_{0}^{L}\frac{dx}{(xg)^{1/2}}$$

3. Sep 16, 2010

### a077456

can you please expain why T = m(X/L)g? Thanks!