# Homework Help: Determing wavelength of sound wave from steel string

1. Feb 7, 2010

### LBRRIT2390

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

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 $$\circ$$C room?

2. Relevant 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 = $$\frac{v}{f_m}$$

3. 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 = $$\frac{1}{2L}$$$$\sqrt{\frac{T_s}{\mu}}$$

2. Feb 7, 2010

### LBRRIT2390

Speed of wave in string: v = $$\sqrt{\frac{T_s}{\mu}}$$
Ts = 100N​
$$\mu$$ = 0.0012kg/m​
v = 288.7​

Fundamental Frequency: F0 = $$\frac{v}{\lambda_0}$$
$$\lambda$$0 = 2L = 2*1.2​
$$\lambda$$0 = 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