Question

and is under 3. A steel- string acoustic guitar has linear density of 5g/m tension of 180 N. The sto is oscillating wave pattern shown . If fixed apart, calculate the frequency of traveling waves (is pt) ia the standing D=75

Answer 1

Given:

$\mu = 5g/m = 5*10^{-3}kg/m$

D = 75 cm = 0.75 m

T = 180 N

Now, the speed of a wave in a stretched string is given by $v = \sqrt{\frac{T}{\mu}}$ .

From the diagram we can write the wavelength of the wave in terms of string length D.

D = 30
[ for each loop the length is $\frac{\lambda}{2}$ ]

.

2D 3

Frequency of wave is given by frequency = wave speed / wavelength.

Therefore, frequency of the wave = $\frac{v}{\lambda}$

$=\frac{\sqrt{\frac{T}{\mu}}}{\frac{2D}{3}} = \frac{3}{2D}*\sqrt{\frac{T}{\mu}}= \frac{3}{2*0.75}*\sqrt{\frac{180}{5*10^{-3}}}=379.47Hz$ [answer]