# Composite wave

1. Jul 25, 2014

### Firben

1. The problem statement, all variables and given/known data

The figure below shows a snapshot of a standing wave on a composite string. It is a node in the composite point. If the aluminium string has a diameter of 1.50 mm, which diameter does the steel string have ?

http://s716.photobucket.com/user/Pitoraq/media/Fys22222_zps61d0eb53.png.html

2. Relevant equations

V = √(F/ρ(length)) = √(F/ρA)

3. The attempt at a solution

Al: Node in both endpoints

y = f(ωt)*sin(kx) <==> sin(k1L1) = 0 <==>

2π/λ1*L1 = P1π <==>

λ1 = 2/p1*L1

Steel: Node in both endpoints

sin(K2L2) = 0 <==> 2π/λ2*L2 = P2π <==>

λ2 = 2/p2*L2

λ = v/f

λ1 = v/f1 = 2/p1*L1 <==> vp1/2L1 = f1 <==>

vp2/2L2 = f2

Same frequency

vp2/2L2 = vp1/2L1 <==>

√(F/ρ(al)A1)*P1/2L1 = √(F/ρ(Steel)A2)*P2/2L2 <==>

√(F/ρ(al)4πd1)*P1/2L1 = √(F/ρ(Steel)4πd2)*P2/2L2 , breaking out d2 <==>

d2 = ((p2)^2*(L1)^2*(ρ(Al))*d1)/((p1)^2*(L2)^2*ρ(Steel))

Where d1 = 0.0015
ρ(Al) = 2.7+ * 10^3/m^3
ρ(Steel) = 7.87*10^3/m^3
L1 = 0.3 m
L2 = 0.25 m

p2 = 5
p1 = 3

Are the nodes from the figure

when i put does values into the equation above i got 2.058 * 10^-3 m

In the answer it should be 1.76 mm

2. Jul 25, 2014

### olivermsun

I think if you are more careful writing the equations (especially the grouping under the square roots) you'll can see more clearly where the problem is. I'll rewrite some of it

Starting from:
\begin{align} \frac{vp_2}{L_2} &= \frac{vp_1}{L_1} \\ \frac{p_2}{L_2} \sqrt{\frac{F}{\mu_2}} &= \frac{p_1}{L_1} \sqrt{\frac{F}{\mu_1}} \\ \frac{p_2}{L_2} \sqrt{\frac{F}{\pi(d_2/2)^2\rho_\mathrm{Steel}}} &= \frac{p_1}{L_1} \sqrt{\frac{F}{\pi(d_1/2)^2\rho_\mathrm{Al}}} \\ \frac{p_2/L_2}{p_1/L_1} \sqrt{\frac{\rho_\mathrm{Al}}{\rho_\mathrm{Steel}}} &= \frac{d_2}{d_1} \end{align}

So it seems as if you forgot d1, d2 were inside the √ and as a result your p1, L1, p2, L2 got squared by accident.

Last edited: Jul 25, 2014