Determing wavelength of sound wave from steel string

[LBRRIT2390]

Homework Statement

A 120 cm-long steel string with a linear density of 1.2 g/m is under 100 N tension. It is plucked and vibrates at its fundamental frequency.

What is the wavelength of the sound wave that reaches your ear in a 20 \circC room?

Homework Equations

Fundamental Frequency

f₁ = \frac{v}{2L}

Fundamental Frequency of a stretched string

f₁ = \frac{1}{2L}\sqrt{\frac{T_s}{\mu}}

Wavelengths of standing wave modes

\lambda_m = \frac{2L}{m}

\lambda_m = \frac{v}{f_m}

The Attempt at a Solution

I solved fundamental frequency as 143.3 then used

\lambda_m = \frac{v}{f_m} to find the wavelength.

I also tried solving for fundamental frequency using

f₁ = \frac{1}{2L}\sqrt{\frac{T_s}{\mu}}

All of my answers have been incorrect. Please help!

 

[LBRRIT2390]

Speed of wave in string: v = \sqrt{\frac{T_s}{\mu}}

T_s = 100N​

\mu = 0.0012kg/m​

v = 288.7​

Fundamental Frequency: F₀ = \frac{v}{\lambda_0}

\lambda₀ = 2L = 2*1.2​

\lambda₀ = 2.4​

f₀ = \frac{\sqrt{\frac{T_s}{\mu}}}{2L}

In the air:

\lambda = \frac{v}{f_0}

\lambda = \frac{v}{\frac{\sqrt{\frac{T_s}{\mu}}}{2L}}

\lambda = \frac{v2L}{\frac{\sqrt{T_s}}{\mu}}

\lambda = \frac{344m/s * 2.4}{288.7} = 2.86