How Does Cross Sectional Area Affect String Wavelength Calculation?

[buffgilville]

1) A cable made from a metal of density 8850 kG/cubic meter, and whose cross sectional area is 9.6 cubic mm is pulled to a stress of 103.4 Newtons. One end of the string is oscillated with a frequency 480.1 Hertz. The wavelength of the wave on the string (in meters) is

T=f^2 * linear mass density * wavelength^2
T=103.4 Newtons
linear mass density = 8850 kG/cubic meter
f=480.1 Hertz
so, I plug all those in and got 2.25E-4
but the correct answer is 7.27E-2

The question gave a cross sectional area. How does that relate to the wavelength?

 

[Andrew Mason]

1) A cable made from a metal of density 8850 kG/cubic meter, and whose cross sectional area is 9.6 cubic mm is pulled to a stress of 103.4 Newtons. One end of the string is oscillated with a frequency 480.1 Hertz. The wavelength of the wave on the string (in meters) is

T=f^2 * linear mass density * wavelength^2
T=103.4 Newtons
linear mass density = 8850 kG/cubic meter
f=480.1 Hertz
so, I plug all those in and got 2.25E-4
but the correct answer is 7.27E-2

The question gave a cross sectional area. How does that relate to the wavelength?

I assume the cross-sectional area is 9.6 square mm, not cubic. The area is needed to calculated the mass per unit length of the string.

The wave equation is given by:
\lambda \nu = \sqrt {\frac{T}{M / L}} where M/L = mass per unit length which is density x area

\lambda = \frac{\sqrt{T/ \rho A}}{\nu}

\lambda = \sqrt{103.4/8850 \times 9.6 \times 10^{-6}}/480.1
\lambda = \sqrt{103.4/.08496}/480.1
\lambda = 34.886/480.1

\lambda = .0727 m

AM

 

[buffgilville]

Thanks Andrew Mason!