Determing wavelength of sound wave from steel string

AI Thread Summary
The discussion focuses on calculating the wavelength of a sound wave produced by a 120 cm-long steel string under 100 N tension. The fundamental frequency is determined using the formula f1 = (1/2L)√(Ts/μ), leading to a speed of the wave in the string calculated as v = 288.7 m/s. The wavelength in the air is found using the relationship λ = v/f0, resulting in a final wavelength of approximately 2.86 meters. The calculations involve understanding the relationships between tension, linear density, and wave properties. Accurate application of these formulas is essential for determining the correct wavelength.
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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

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

\lambdam = \frac{v}{f_m}

The Attempt at a Solution



I solved fundamental frequency as 143.3 then used

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