Determining Steel String Diameter for Standing Wave Node

[Firben]

Homework Statement

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

Homework Equations

V = √(F/ρ(length)) = √(F/ρA)

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

 

[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:

Same frequency

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

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

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

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.