String 1 has linear density 2.0g/m and string 2 has linear density 4.0g/m . A student sends pulses in both directions by quickly pulling up on the knot, then releasing it. Consider the pulses are to reach the ends of the strings simultaneously.

What's the Length of string 1 and string 2? ( sum of L_1+L_2=4m)

## Answers (1)

Let be x = L_1, y = L_2

The wave speed takes the general form

v = SquareRoot( B / d)

where B is the elastic property (bulk modulus) and d the inertial property (density).

So we have

x = vx * T = K * SquareRoot(B/2) * T

y = vy * T = K * SquareRoot(B/4) * T

where T is the time to reach the ends of the string (same for both)

Now we have the equations system

x = y * SquareRoot (1/2) / SquareRoot (1/4) =

= y * 2 /SquareRoot(2) = y * SquareRoot(2)

x + y = 4

Solving: x + y * SquareRoot(2) = 4 --->

y = 4 / [1 + SquareRoot(2)]

x = 4 SquareRoot (2) / [1 + SquareRoot(2)]

Solution:

x= 4 √2 / (1 + √2)

y = 4 / (1 +√2),

Note: We can use the linear density instead of the volumetric one because the strings sections are supposed to have the same value.

I hope it help.

Reference:hyperphysics.phy-astr.gsu.edu/hbas...